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Permutation and Combinationby Total Gadha - Sunday, 17 August 2008, 11:13 AM

Dagny has been telling me smugly that it's now my turn to write an article. So I decided to write one assoon as possible and pass the buck back to her. We both know that it takes a lot of pain to research foran article and write the text so we both try to avoid it for as long as possible. On the other hand, sincewe both enjoy writing and like to publish a well-finished chapter, we both undertake the task eagerlyonce the ball is in our court. Writing these chapters have made me realize how true is the maxim "if youwant to learn something well, teach it." I can cite a lot of topics which I learnt while teaching in the classor answering students' questions. These years of teaching have made so many topics seem prettychildish to me which were nightmares to me before. But knowing a topic well and teaching it are twodifferent ball games. Many a times, I have found myself struggling in the class to make a studentunderstand something which I have found obvious to understand! The fault is not on the student's part.Three or four years ago, I would have been stuck on the same point! Every time I encounter a situationlike this, my only measure is to stop at that point, retrace my steps with the students and slowly unravelthe difficulty he's facing. It works most of the times. But sometimes, the student is in so much awe withthe topic that his mind gets frozen. I have seen this happening many a times. Topics such as time, speed

and distance, Permutation and Combination, geometry etc. inspire so much fear among the students that sometimes simple principles,which they would have otherwise understood do not strike them as simple. I get incessant queries such as "Sir, time speed distancechhod sakte hain?," "Sir, permutation combination na karein to chalega?" from my students all the time. And the sad part is, that thelevel of CAT in these topics is very simple. They are doable. In the present chapter, I am trying to present one of these dreadedchapters as I understand it. I hope my students understand it too.



I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on thisin the CBT Club this week.

If you think this article was useful, help others by sharing it with your friends!

You might also like:ProbabilityMaxima, Minima and Inequalities

Reply

Re: Permutation and Combinationby vamsi krishna - Sunday, 17 August 2008, 12:23 PM



Sir,

Tatz a GEM OF AN ARTICLE

thnkz,

VaMsI

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Re: Permutation and Combinationby Amit A - Sunday, 17 August 2008, 05:22 PM

Now what do i say other than "THANKS A LOT TG !!! "

u rock ....

Amit

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Re: Permutation and Combinationby Amit Kumar - Sunday, 17 August 2008, 06:14 PM

Very Very useful article..

Thanks

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Re: Permutation and Combinationby Leonidas S - Sunday, 17 August 2008, 06:20 PM

Hi TG,

A fabulous article.I think there is a m istake in the solution of Prob 3 of the 4 friends problem.

It says "The total number of ways in which 4 friends can stay together is

104 " I think it should be 10000 as you found out earlier.

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Re: Permutation and Combinationby Amit Kumar - Sunday, 17 August 2008, 09:40 PM

Hi TG,

Answer to 'Good Boy Bad Boy' movie question:

Since these two couples want to sit together at e ither end. Now, we'll tie them together. Hence, total number of students will be 7 out ofwhich 6 students can sit in 6x5x4x3x2x1 = 720 ways. and the couple can sit at e ither ends (i.e 2 ways) and the couples can sit in 2 waysamong themselves and finally, in each couple, boy and girl can sit in 2 ways(BG and GB)(Total = 2x2 ways). Hence, Number of ways all cansit will be:

720x2x2x2x2 = 11520 (Answer)

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Re: Permutation and Combinationby Manika Tandon - Sunday, 17 August 2008, 10:41 PM

hey TG...

awesome piece of writing dude... rather thanx dagny for giving him a push to write it... :D

both of u a ll rock..

cheers,

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Re: Permutation and Combinationby AsHwIn Drmz - Sunday, 17 August 2008, 11:02 PM

hi....Tg....



G8 article...thk u so much for a wonderful article,actually i was worried as u were not writing articles for cat 08 students....but nw m so glad

and confident dat my all worries will vanish.Tg can u plz write an article on probability too....

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Re: Permutation and Combinationby Ashwin A - Monday, 18 August 2008, 07:51 AM

TG Sir!Simply Fantabulous!! Just wanted to know do u have a Book having all topics of Quant...I have the book on Number theory which in itse lf,

is very good.!!!

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Re: Permutation and Combinationby Dagny Taggart - Monday, 18 August 2008, 09:17 AM

Hey Manika,

TG played smart this time. I asked him to write an article three days back ;he wrote, edited and uploaded it within two days. Now the ball is

back in my court.

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Re: Permutation and Combinationby Diwakar Bhaskar - Monday, 18 August 2008, 09:47 AM

Thnks a lot TG...........i dont have word to say anything more ...................

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Re: Permutation and Combinationby vamsi krishna - Monday, 18 August 2008, 11:13 AM

Hey Amit,

Pl Recheck

Shudn't Ans be 720 x 2 x 2 x 2

1..six students can be arranged in 6! = 720

2..The 2 couples can sit at e ither of the ends in 2 ways

3..and then each couple on e ither side can exchange their seats..in 2 ways so two couples can do tat in 2 X 2 ways

regards,

VaMsI

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Re: Permutation and Combinationby Krushang Shah - Monday, 18 August 2008, 11:36 AM

THANKS A LOT TG!!!its a wonderful article.....

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Re: Permutation and Combinationby amit kumar - Monday, 18 August 2008, 01:44 PM

Thanks TG

Awesome Article.......

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Re: Permutation and Combinationby nikhil khandelwal - Monday, 18 August 2008, 04:58 PM

Thanks a lot TG

Doing a great job



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Re: Permutation and Combinationby rajdeep choudhuri - Monday, 18 August 2008, 05:46 PM

Hi TG,

The article is simply m ind blowing!!!Plzzzzzzzz post an exercise on Permutation and Combination problems.

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Re: Permutation and Combinationby bryan cole - Tuesday, 19 August 2008, 12:59 AM

Is the answer to a+2b+3c.....total number of e lements is 36?

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Re: Permutation and Combinationby Total Gadha - Tuesday, 19 August 2008, 03:37 AM

Hi Rajdeep,

About the exercise, have you seen the article complete ly?

Total Gadha

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Re: Permutation and Combinationby rajdeep choudhuri - Tuesday, 19 August 2008, 04:22 PM

Yes .But i fee l it will be lot more helpful if you can give us an exercise like the one on Number System problems,I mean plenty moreproblems to solve.Also where can i get the answers to these problems???

Anywys TG rocks!!!!

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Re: Permutation and Combinationby Riyaz Iqbal - Wednesday, 20 August 2008, 11:03 PM

Hi TG,

Excellent article,as usual.I've a question about the 'counting functions' problem.The no.of functions from a 5-e lement set to 7-e lementset should be less than 7^5,right? 7^5 simply represents the number of ordered pairs (x,y) from first set to the second.Several of themmay correspond to a single function.eg: (1,1) (2,2),...(5,5) are five distinct ordered pairs but a single function viz y=x. Am I m issingsomething?

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Re: Permutation and Combinationby ands ands - Thursday, 21 August 2008, 07:22 AM

ash, hi I've just started with the Num system but am finding the area vast,enomously huge and the topics not concentrated from aparticular source.From your post I see that there is some consolidated resource for the Num system you ta lk in about or is the the ebookby TG ppl.Can you post on the name here.It would be sure help if I am to find the book and the contents satisfying.Thanks.....!!!!

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Re: Permutation and Combinationby Total Gadha - Thursday, 21 August 2008, 01:18 PM

Hi R iyaz,

There are 75 arrangements, and every arrangement, mapping 5 e lements of set 1 to 7 e lements of set 2 would be counted as a singlefunction only.

Total Gadha

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Re: Permutation and Combinationby Shalini B - Thursday, 21 August 2008, 02:45 PM

Hi TG,



I have registered for your Copycat exams by paying through credit card but I am facing problems while trying to write the exam online. Ihave mailed you long back about this (to the copycat email ID and also to admin email ID). But I have not received any answer for that.And I could not find any contact number too on this site where I could call up and report my problem. Please respond.

Content from my previous mail dated July 30th 2008 to you below:

My problem is that when I tried tak ing copycat 2 the site was unresponsive and did not navigate to the next page at a ll. I tried tak ing theexam but when I started with it and entered my options for the first page and clicked on the Next button, the site became very slow anddid not show the next page at a ll. I tried several times but it was of no use. I checked the browser compatibility test and it said my systemis fine for tak ing the test and my internet connection is good and all other websites are work ing fine. I am mighty disappointed. Pleaselook into this problem ASAP

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Re: Permutation and Combinationby manoj 1123 - Thursday, 21 August 2008, 04:06 PM

Hi TG,Though I fee l not ok to write here, I am also facing the same problem, not able to browse to next page. may be if the page remains openfor some time the connection to ur server breaks. P lease send the pdf for the COPYCATs to my mail beforehand.So that I can take print and appear the test.. P lz find a solution for us.

Thanks.

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Re: Permutation and Combinationby Leonidas S - Friday, 22 August 2008, 10:33 PM

Hi TG,

That was an excellent article

Can you please post some articles on functions,logarithms and probability

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Re: Permutation and Combinationby rakesh ojha - Sunday, 24 August 2008, 08:02 PM

hi dear

First of a ll ..thnx a lot for such a nice collections of questions which has helped to build the fundamentals of this topic...

i just have one doubt in one of the examples ..above where we are getting number of 7 didgit numbers ...in binary.....that part is ok...

but when the number is converted to decimal base....there will be other conditions aprt from last 3 digits as zeros........say a binary nowhen conveted to decimal and becomes 24...that is divisble by 8....

may be i m wrong somewhere but plz te ll me why only you have taken the condition of last 3 didgits as 0 for dibisiblity by 8...why not otherconditions....

thnx

rakesh

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Re: Permutation and Combinationby Total Gadha - Sunday, 24 August 2008, 10:12 PM

Hi Rakesh,

Convert 24 to binary please?

Total Gadha

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Re: Permutation and Combinationby amit kumar - Monday, 25 August 2008, 01:31 AM

HI ALL,question.>there are 10 students out of which three are boys and seven are girls,in how many different ways can the students be pairedsuch that no pair consists of two boys?

Unable to solve this.Please help wid proper explanation

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Re: Permutation and Combinationby IRONMAN A - Monday, 25 August 2008, 11:57 AM

Thanks a m illion for the beautiful article TG.

I have a small doubt. P lz do expla in me.

"There are 6 tasks & 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned either to person 3 orperson 4. Every person must be assigned one task. In how many ways can the assignment be done(CAT 2006)?"

What's the wrong in my procedure....

First person: As task 1 & task 2 cannot be assigned any of the other 4 tasks can be assigned to him - 4 waysSecond Person: As task 1 & task 2 cannot be assigned, and one already assigend to Person 1, any of the other 3 tasks can be assignedto him - 3 waysThird Person & Fourth person: one of these get Task 2. The other gets any of the other 3. (one of tasks 1,3,4,5,6 (leaving the tasksassigned to Persons 1 & 2) - 3 waysFourth & Fifth Persons: other 2 tasks can be assigned in 2 ways.

total = 4*3*3*2 = 72

Plz do answer me.. thanks in advance

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Re: Permutation and Combinationby arpi sin - Monday, 25 August 2008, 12:20 PM

TG/Dagny Wonderful work!!!I was going through the number system quiz,and have a doubt in a question .Where can I post my query ?

RegardsArpita

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Re: Permutation and Combinationby Surendran Chandravathanan - Monday, 25 August 2008, 09:43 PM

Hi Amit,

Here 's my try...

1st Approach

1 boy can pair up wid 7 girls in 7 ways...So for 3 boys, total no. of ways is 21..And 1 girl can pair up wid 6 other girls in n(n-1)/2,ie, (7x6)/2 = 21 ways..(or) Just 7C2 = 21 will cede u the answer.

So the final ans is 21 + 21 = 42 ways..

[[OR]]

2nd Approach

Should this problem be solved like below....???

To se lect 2 ppl from 10, we can do in 10C2 = 45 ways..And arranging them in pairs can be done in 45x2 = 90 ways..(or) juz we have toarrange 2 places out of 10 which can be done in 10P2 = 90 ways..

And then, 3 boys can be paired up in 6 ways..

So is the final ans 90 - 6 = 84 ways ??????

I am perplexed..

Which ans is correct, TG ????? Appreciate ur comments.........

Best Rgds,

Suren

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Re: Permutation and Combinationby Surendran Chandravathanan - Monday, 25 August 2008, 09:58 PM

TGGGGGG,

Ur art of solving & simplifying the problems leaves me in awe but it invites more confusions.....

I wud say dat the below problem can be solved easily but u said it is not so..

In how many ways can four persons be seated out of 5 boys and 3 girls on four different seats?

Now see, totally thr are 8 ppl where they 've to be seated in 4 seats..

So won't the normal permutation formula do ???????



ie , 8P4 = 1680 ways....

P lz e lucidate the fact for break ing the problem & considering 4 cases in ur approach....

Thanks in advance,

Best Rgds,

Suren

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Re: Permutation and Combinationby Black i - Monday, 25 August 2008, 11:38 PM

Hi TG,It is regarding the '7 digit binary no div. by 8 when converted to decimal no' question.How can one decide that the last 4 bits , if are equal to zero then the no is divisible by 8 when converted to decimal ?Suppose the same question asks about octal no instead of binary, how should one decide about it ?

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Re: Permutation and Combinationby amit kumar - Tuesday, 26 August 2008, 10:21 PM

Hi Answer is 630.give another try.

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Re: Permutation and Combinationby Harish Bansal - Thursday, 28 August 2008, 12:51 PM

Hi TG,Gr8 article....Plz post the solutions of the problems that you have mentioned in ur artical...We want to match our answers and approach.And plz can u post something on probability prior to CAT 2008.

Eagrly waiting for a reply.........

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Re: Permutation and Combinationby Tuhin Banerjee - Tuesday, 2 September 2008, 05:14 PM

Hi TG, it's realy a mind blowing article on permutation and combination.

I have question on the same topic and have two different approaches , kindly tell me which one correct :

Question: Ten students have been shortlisted to form two teams of six students each, such that there are exactly three commonbetween the two teams, In how many ways can the teams be formed.

Ans:

1st approach:

Three common stuednts we can take out of 10 students = 10 C3 ways ,

We need to build each group having 6 students , for the 1stGR. rest of the 3 students we can take out of 7 students =7C3 ways

So , 1stGR complete, for the 2ndGR we can take rest of the 3 students out of the 4 students in = 4C3 ways.

Total = 10 C3 *

7C3 * 4C3 = 16800

2nd Approach :

From 10 students , 3 students commonbetween the two teams can be selected in = 10C3 ways.

since 7 students are still left and for each team we need 6 students , Now select the student

will find a place in neither team and it can be done in = 7C1 ways.



Now remaining 6 students have to divided into two groups of 3 each and it can be done in

= 6C3 / 2

Now two teams can be formed = 10C3 * 7C1 *

6C3 / 2 = 8400

please tell me where I am missing the concepts

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Re: Permutation and Combinationby rudra samal - Tuesday, 2 September 2008, 10:01 PM

Hi TG,

In the example where u asked:

How many arrangements r dere of six 0's, five 1's and four 2's, where

(i)first 0 precedes first 1?

shudnt the ans be 343980 and not 349380?.... and there 's an easier way as well. It goes like this:

total arrangements=(15!/(4!*5!*6!))

so ans will be =total arrangements*(p/q)

where q=all possibl arrngmnts of the 1s and 0s

p=arrngmnts where 0 occupies 1st position

so,

q=11!/(6!*5!)

p=10!/(5!*5!)

ans=343980

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Re: Permutation and Combinationby Total Gadha - Wednesday, 3 September 2008, 10:49 AM

Hi Rudra,

Corrected the typing error. And good way to solve the question.

Total Gadha

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Re: Permutation and Combinationby Total Gadha - Wednesday, 3 September 2008, 10:51 AM

Hi Tuhin,

The second solution is correct. First has repetition. P lease check the question of "three states with three students from each state...."

Total Gadha

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Re: Permutation and Combinationby bhageerath n - Wednesday, 3 September 2008, 10:30 PM

hii'm here for the first time.. this was pretty helpful.. how do i cross check if i got the right answers?? will the solutions be available for thequestions??

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Re: Permutation and Combinationby bhageerath n - Wednesday, 3 September 2008, 11:04 PM

Hi TG,for the question -In how many ways can four persons be seated out of five boys and three girls on 4 different seats?

should we take those cases as mentioned by you??why cant we do it like this?



there are 8 people in a ll.. se lect four of them in 8c4 ways (70 ways). Now you can arrange these four in 4! ways(24 ways). So total ways is70*24 = 1680.

is there anything wrong with this approach??

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Re: Permutation and Combinationby Syd S - Thursday, 4 September 2008, 01:58 PM

kudos for another brilliantly structured and edited article,,,!!!!

was just wondering guys bout dis prob,,,"in how many diff ways can the faces of a cube be painted in 6 diff colors?"

shudnt the no of ways be 6 X 5 x 3!,,as the first face can be painted in 6 ways,,rather than only 1 way,,

jus wanted 2 knw,,y is it 5 X 3! rather than 6 X 5 x 3!

thnx

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Re: Permutation and Combinationby brijesh gogia - Saturday, 6 September 2008, 07:12 PM

Thank You for this very useful article.

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Re: Permutation and Combinationby ganesh p - Sunday, 7 September 2008, 08:41 AM

That U Very much sir ..........as expected my search for the best article on permutation and combination did end here..

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Re: Permutation and Combinationby k ladad asda - Sunday, 7 September 2008, 08:44 PM

just being curious. are u vamsee krishna from CSE IITKGP, class 2006??

If you are k indly reply back, it would be nice to know one more familiar face at TG

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Re: Permutation and Combinationby Manish Kashyap - Tuesday, 9 September 2008, 07:56 PM

Hi TG,

My Solutions are -

1)a

2)c

3)b

4)

5)c

6)

7)c

8)c

9)a

10)b

11)b

12)c

13)b

14)b



15)

16)

17)b

19)d

20)

21)c

22)1672523)2424)840

25)120

26) 68027)26!/228)45

29)21

30)3531)532)7c233)33600I was not able to solve some questions for which I had not put any answers. Can you please verify my answers andtell me the solutions of which are wrong or which I was not able to solve.

Thanks and Regards,

Manish

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Re: Permutation and Combinationby Ankit Arora - Wednesday, 10 September 2008, 05:27 PM

Hi TG,

I haven't solved all the questions till now...just want to confirm the answers of first 5 ques..

1 (a)

2(b)

3(d)

4. I m getting (26^3)*220 as my answer

5(c)

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Re: Permutation and Combinationby Manish Kashyap - Thursday, 11 September 2008, 03:09 PM

Hi TG Sir,

P lease check the solutions pleazzzzzzzzzzzzzzz. I am waiting for solutions.

Thanks and Regards,

Manish

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Re: Permutation and Combinationby vamsi krishna - Friday, 12 September 2008, 04:25 PM

Frns



Check/discuss answers for above problems @

http://totalgadha.com/mod/forum/discuss.php?d=3683

VaMsI

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Re: Permutation and Combinationby Saurabh Shekhar - Friday, 26 September 2008, 09:59 AM

Hi TG,

Could you please post the complete solution to the Exercise.

We are all eagerly waiting for it.

Saurabh

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Re: Permutation and Combinationby Venkkatesan R - Saturday, 27 September 2008, 04:44 AM

Hi suren,

I think the question involves just the se lection. Hence the arrangement is not required. Hence 10c2-3c2=45-3=42 is the answer.

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Re: Permutation and Combinationby arvind jenamani - Saturday, 4 October 2008, 09:19 PM

Hi TG sir,

The link is fabulous. In combination part there is a question " In how many ways 5 boys , 3 girls can be sitted in 4 seats?"

There we don't need to segregate into parts like you mentioned.

8C4 ways we can se lect 4 people(without considering boy,girl combination)

4! ways to arrange them. so, 8C4*4!=8P4=1680

which match also your answer.

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Re: Permutation and Combinationby Amit Trivedi - Saturday, 11 October 2008, 10:31 PM

Hi TG sir,

The lesson is really great. i have a doubt for one of the examples.

the examplein which 5 boys and 5girls go to watch the cinema spiderman3.

you gave two arrangement for the wish of the boys "every boy wants to sit with a girl" and no two girls sit together

i think other arrangement of the form are possible

G B B G B G B G B G

G B G B B G B G B G

G B G B G B B G B G

G B G B G B G B B G

In the above arrangements every boy is sitting with a girl and no two girls are sitting together

clarify my doubt, let me know where i am making mistake in reasoning.

Warm Regards,

Amit

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Re: Permutation and Combinationby Amit Trivedi - Saturday, 11 October 2008, 11:57 PM

if the couples are siting at the ends

the arangements are 2 X 2 X 2 x 6!

Amit

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Re: Permutation and Combinationby meena k - Saturday, 18 October 2008, 04:24 AM

hi,

for the states question can't we solve l ike this from 3states 3 students eachso selecting 5 frm 9= 9c5

then subtracting 6c5 (case where al l 5 students are selected frm two states)

ie:9c5 - 6c5 .pls tel l me where the problm is

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Re: Permutation and Combinationby noel sandesh - Monday, 3 November 2008, 03:40 PM

Hi TG,

This is in regard to an example problem (No. of 7 digit binary numbers that are divisible by 8 when converted to base 10). i was wonderingas to y the 3rd digit from the last cannot be 1 or 0. You've mentioned that the last 3 digits should be 0. But if the last 3 digits are 100then its base 10 equivalent will be 400 which is divisible by 8.... I'm really bad at P n C but i can't figure out the logical flaw in mythought... Can u help me out on this?

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Re: Permutation and Combinationby rajesh kambampati - Thursday, 6 November 2008, 12:36 AM

Ya i agree with you sandesh, i too got this doubt.And i think we are correct.

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Re: Permutation and Combinationby Total Gadha - Friday, 7 November 2008, 12:49 AM

(100)2 = 4 (in decimal)

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Re: Permutation and Combinationby Complete Gadha - Friday, 7 November 2008, 08:03 PM

Hi TG Sir,

This is the best material i have ever come across for Permutation and Combination.

I have been going through the material in the forum. all of them r so amazing that i am actually gaining the confidence to crackthe CAT though i have not been doing so good in the mock cat. Thanks a lot..!!

Hats off To U...!!!

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Re: Permutation and Combinationby Bheemesh K - Wednesday, 12 November 2008, 01:40 AM

Hi TG,

I had a doubt in the first problem (i.e 10 speakers problem) solved in the permutation section.

I solved it in this manner. you tie the PM, MP and MLA into a single person and then calculate the ways to arrange them. wudn`t it be 8! then that we have other 7 speakers and 1 group of PM, MP and MLA to arrange?

I find your way also logical but am not able to understand difference between my approach and your approach. k indly can you expla inwhere I am missing the logic?

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Re: Permutation and Combinationby sumit jamwal - Wednesday, 4 February 2009, 08:51 AM

Hi TG,

I have some problem in understanding the concept of partition problem..

of 5 balls as stated be you.

let's take * * * * * these as 5 sim ilar balls

now dividing them into 3 grps..

i can do it by placing 2 sticks ..for that i have total of

|*|*|*|*|*| --> | denotes choices

6 choices..

so se lecting 2 spaces out of 6 can be done in

6C2 = 15 ways..

te l me wats goin wrong in this approach..

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Re: Permutation and Combinationby sumit jamwal - Thursday, 12 February 2009, 09:41 PM

hi TG,the problem i had above , i'd done a mistake actually thr wud be 30 ways ...

for 1st stick - 6 waysfor 2nd stick - 5 so total - 6*5=30.

if you cud clarify this thing it wud be nice

Regards,sumit

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Re: Permutation and Combinationby Total Gadha - Thursday, 12 February 2009, 11:09 PM

Hi Sumit,

In your solution, you can't place sticks like this for example||* * * * *

This would denote a 0 0 5 case which you cannot cover. You can only cover 0 5 0 case by placing the sticks in this manner: |* * * * *|

The way you are placing the sticks you can never place two sticks side by side because you are tak ing it to be one space only.

Total Gadha

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Re: Permutation and Combinationby sumit jamwal - Friday, 13 February 2009, 12:16 AM

hi sir,sir that's fine ..bt if iam nt considering this case ..then y is my answer comin to be arnd 30 ..whr as u stated it's 21.

i think thr m ight be somekind of repeatition...

and as u told | | * * * * * nt included so,

now suppose i place 1st stick then arrangement wud be like..| * * * * *

now to include ur case for 2nd stick thr are 6 cases as i can placestick between stars and beside 1st stick

in this case total cases come out to be 36 still more than 21

a bit confused as how to get correct result by using this stick method.



regards,sumit

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Re: Permutation and Combinationby Total Gadha - Friday, 13 February 2009, 09:32 AM

Hi Sumit,

The repetition is because you are tak ing partitions to be different whereas there have to be taken as sim ilar. For example, take two simplecases:

* P1 * * * P2 * and * P2 * * * P1 * where P1 and P2 denote the partitions. Both these case would yie ld the same result, i.e. 1 3 1 but you

are counting them as different.

If I do this by your method I will do it like this:For the first partition I have 6 places. * * | * * *

For the second partition I have 7 places. Therefore total number of ways = 6 7 = 42. As each case is be ing repeated once, the totalnumber of cases = 42/2 = 21.

Total Gadha

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Re: Permutation and Combinationby sumit jamwal - Friday, 13 February 2009, 05:07 PM

hi TG,

thanks ,that was wat i think ing....

gr8 work ..

Regards ,

sumit

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Re: Permutation and Combinationby Abhinav Kumar - Friday, 22 May 2009, 01:38 PM

Thats really a beautiful one.

Made such a difficult task so simple with the examples of such a variety.

Thanks a Lot

Abnv

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Re: Permutation and Combinationby rony roy - Friday, 22 May 2009, 02:33 PM

hi TG,

what was the answer to the question that goes.....

how many arrangements of six 0s five 1s and four 2s are there in which

ii)the first 0 precedes the first 1,(and) precedes the first 2.

my answer comes out to be a massive : 504504

reply soon...

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Re: Permutation and Combinationby iim freak - Tuesday, 26 May 2009, 12:29 PM



hi tg.. 1st of a ll thanks a ton for this simply awesome post .. i ws nt getting any gud materia l on P&C n Probability n was worried ... thanksagain..

.. regarding the Problem :: there are 10 speakers who r supposed 2 address. One constraint is PM shud address before MP and MP beforeMLA .

My soln:: Let PM = P ; MP = Q; MLA = R

the order shud be PQR .. so i tuk PQR as 1 block and so total people 2 b arranged becomes (7+1) = 8 ... and now arranging them we get

Ans = 8! .

P lease te ll where am i making the mistake ?

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Re: Permutation and Combinationby Priyesh Tungare - Wednesday, 27 May 2009, 02:59 PM

Hi TG,

Thanks for the article.

I have one doubt.

In the question, How many 3-digit numbers are even and with no repeated digits, in the second case for number not ending in 0, you havemultiplied 8*8*4.

Can you expla in why we are multiplying 4 here?

This may be simple question for others, bt i be lieve in saying.. when in doubt ask....

Thanks,Priyesh Tungare

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Re: Permutation and Combinationby Total Gadha - Thursday, 28 May 2009, 02:06 PM

Hi Randeep,

'Before ' m ight not a lways mean 'just before,' as you have assumed by tak ing them together. There can be other persons between P, Q andR.

Total Gadha

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Re: Permutation and Combinationby Total Gadha - Thursday, 28 May 2009, 02:08 PM

Hi Priyesh,

There are four 4 even digits after removing zero- 2, 4, 6 and 8. The number would end in one of these.

Total Gadha

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Re: Permutation and Combinationby iim freak - Thursday, 28 May 2009, 10:21 PM

oh yeah ,, gosh i shud hav thot dat.. thanks a lot for clearing the doubt ...

TG sir it will b a gr8 help if u cud post some gud text, like dis, for Probablity a lso .. dats another area, which i find tough..

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Re: Permutation and Combinationby iim freak - Friday, 29 May 2009, 10:09 AM

hi tg.. from where can i find the answers 2 the above exercise.. its v imp 2 chk whether i hav understud ur concepts pr nt.

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Re: Permutation and Combination



by anirban bhar - Sunday, 31 May 2009, 12:34 PM

hi tg,im a bit confsed wid the solution to the paint cube problem............

the firstface wich we choose cn b painted in 6 ways,.........y havent u considered that....plzzz enlighten

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Re: Permutation and Combinationby Avishek Chakraborty - Thursday, 11 June 2009, 07:39 PM

Hey guys I have a question re lated to this topic, could u solve this? Here it goes...

Q. 4 boyz n 4 girls are arranged in a row so that no 2 girls are together.Wat is the probability that any 2 boyz are together?

waiting for ur response...

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Re: Permutation and Combinationby iim freak - Thursday, 11 June 2009, 08:27 PM

dude .. this is PnC thread n u ask ing Probability ..jst k idding .. anyways i dnt know probability so cant help @ this moment .. TG sir will beposting a text on Prob soon... waiting for that...ATB mate

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Re: Permutation and Combinationby ku k lux k lan - Thursday, 18 June 2009, 07:18 PM

there are 4 friends and 10 hote l roomsq)in how many ways can at least two friends stay together

my answer to this is as follows,which is leading me to the wrong answer,will be very grateful if someone can point out the fa llacy in this

1 case)-2 friends stay togetherc(4,2)*10*9*8

2nd case)three friends stay together

c(4,3)*10*9

3rd case)all four stay together

10

adding up all these cases i am getting 4690,which is wrong....please te ll me the mistake in this

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Re: Permutation and Combinationby rohan kaushal - Saturday, 20 June 2009, 09:03 AM

hi,

Q. how many three digit numbers are there which are even and have no repeated digits ????

can't it be this way???

_ _ _the 3rd place 5 choices ,(i.e; 0,2,4,6,8) so no of ways are 5 1st place have 8 choices (excluding 0 and the 3rd place 's number)& 2nd place have again e ight choices ....and that comes out to be 8 * 8 * 5 = 320 ..

could you plz help me .......thanx a lot!!!!

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Re: Permutation and Combinationby amit ja in - Sunday, 21 June 2009, 12:37 PM

@Rohan,

When you will use 0 for the unit's place then for the hundreds place you will have 9 choices(1....9) and 1oth place 8 choices.

I hope now you know where you are getting it wrong

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Re: Permutation and Combinationby srinivasan ravi - Sunday, 12 July 2009, 10:51 AM

hi sir,can u pls expla in me the problem just before cicular combinations..i cant understand it clearly..pls help..thanks..

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Re: Permutation and Combinationby smruty panigrahi - Monday, 13 July 2009, 11:10 AM

Hii,

This is a rather simple look ing problem but has been pestering me and am unable to solve. P l chech it out and help me guys.

You have to find out the number of ways to reach from A to B.

Extremely sorry 4 the poor quality of fig. I a in't that good with computers.

Plz give a detailed solution.

Thanx in advance.

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Re: Permutation and Combinationby aryan raj - Monday, 13 July 2009, 03:03 PM

hi

sir

this file is so good for cat exam so i want solution of the all prob

so plz sand me the solution on my email

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Re: Permutation and Combinationby Ankit Megotia - Tuesday, 28 July 2009, 11:13 AM

hey smruty,by any chance do you know the answer to your question?



its coming out to be 72.if its correct i will te ll you the procedure.

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Re: Permutation and Combinationby Jaideep Das - Tuesday, 28 July 2009, 10:44 PM

Can somebody provide me the answers for (ii) and (iii) parts of the question for 6 0's, 5 1's ans 4 2's?

I got the answers as 15840 for (ii) and 378738 for (iii)

Are these answers correct? Please reply

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Re: Permutation and Combinationby Gowtham Muthukkumaran Thirunavukkarasu - Monday, 3 August 2009, 11:20 PM

You r right.The solution given here, is based on a assumption that the three speek in order, which is not the case. Your solution is perfect i guess.

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Re: Permutation and Combination to Bheemesh Kby Gowtham Muthukkumaran Thirunavukkarasu - Monday, 3 August 2009, 11:22 PM

You r right.The solution given here, is based on a assumption that the three speek in order, which is not the case. Your solution is perfect i guess.

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Re: Permutation and Combination TO smruty panigrahiby Gowtham Muthukkumaran Thirunavukkarasu - Tuesday, 4 August 2009, 12:31 AM

From the figure you have given i have worked out and got the answer to be 30. Juz check if you already know the answer and if it is wrongplease let me know. I have attached the file here.

Moreover I have assumed that one can travel through a path only once. For example you cannot go through ED and again comeback to E.

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Re: Permutation and Combinationby Avishek Chakraborty - Wednesday, 5 August 2009, 03:31 PM

Firstly i must say the article is awesome!

Please pardon me if my question seems childish!

There 's one question re lated to dice, in how many ways sum of 8 can be obtained by rolling 2 dice if the they are distinguishable...

now my question is as we are rightly considering (3,5) and (5,3) as different cases, why are we not tak ing two cases for (4,4)? Though thedenominations are same but the dice are distinguishable.

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Re: Permutation and Combinationby cat champ - Saturday, 8 August 2009, 04:11 PM

i dont think it will be taken abhishek.. as(4,4) is making the dice indistinguishable..

btw.. very good work tg sir..the best article till date i have found on p&c.. and i will be posting the solns soon..

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Re: Permutation and Combinationby animesh chandan - Saturday, 15 August 2009, 10:57 PM

hi TG i hav problemQ:there are 7 men and 5 women we have to se lect 4 men and 3 women but if mrs A will go then mrs B will not go.in how many ways v canselect them?

my approach is 7 men will be se lected in 7C4 waysand there are 3 casescase 1: we first se lect mrs A them automatically mrs b will not be se lected then we can se lect remaining women in 3C2 ways,that is 7C4*3C2=105.case 2: we se lect mrs B then automatically mrs A will not be se lected then ve can se lect remaining women in 3C2 ways,that is 7C4*3C2=105,

case 3:if both of them are not se lected then wewill have 3C3 ways to se lect womenthat is 7C4*3C3=35 that means total no of ways =105+105+35=245

plZZZ te l me is it the right approach?????

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Re: Permutation and Combinationby Anoop Maithani - Thursday, 20 August 2009, 10:30 AM

Hi TG..

Amazing article ..waiting eagerly for an article on Probability..

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Re: Permutation and Combinationby BellTheCAT ... - Thursday, 20 August 2009, 07:18 PM

Hi TG,

A really helpful article,for it he lped me revise P&C. thanks a lot.

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Re: Permutation and Combinationby deep sin - Saturday, 5 September 2009, 01:59 PM

@ ku k lux k lan

You are forgetting to take one case into calculation

The case is : - 2 stay in same hote l and remaining 2 also stays in same hote l.

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Re: Permutation and Combinationby Pallav Jain - Thursday, 17 September 2009, 07:03 PM

Hi All,

Can you plz te ll me the solution ?

Using one or more of the digits 1, 2, 3, 4, 5, 6 and 7, how many 7-digit numbers can be formed which are divisible by 7?



Options :-

A) 75 + 74 + 73 + 72 + 7 C) 4 76

B) 4 75 D) 76 E) 6 76

Thanks

Pallav Jain

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Re: Permutation and Combinationby priya ramraj - Friday, 18 September 2009, 04:00 PM

Hi TG,I have a doubt in the article that u have posted. To get a sum of 8 with distinguishable dices, wont the occurrence of(4,4) be differentbecause it is distinguishable??? shudnt we consider it as 2 cases ..P ls clarify...

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Re: Permutation and Combinationby ROHIT K - Friday, 18 September 2009, 05:05 PM

Hi priya,

Jus to giv it a try..

To get a sum of 8 with distinguishable dices, wont the occurrence of(4,4) be different because it is distinguishable??? shudnt we consider it as 2 cases..

(2,1) and (1,2) are different in distinguishable dice(which is not the case in same k ind of dice).

Say, you have a red and a blue die

Case: (1,2) and (2,1)

1.Red->1 Blue->2

2.Red->2 Blue->1

You can distinguish the two not only by their color but a lso by the difference in their numerical value.

Case 2: (4,4)

1.Blue->4 Red->42.Red->4 Blue->4

Aren't they both same??

If yes, then we need not consider it two cases.

In short, since the numerical value is same, we cannot distinguish the case though the two dice are distinguishable.

Hope this helps.

Rohit

P.S if you still didn't understand...TG to hai..!!

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Re: Permutation and Combinationby priya ramraj - Friday, 18 September 2009, 05:31 PM

Hi Rohit,going by ur example,Wat i understand from the article is that color is the distinguishing factor and not the number... If we remove the color aspect from the argument only then the distinguishingproperty vanishes... hope i am not wrong..anyways.. Thanks for the approach

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Re: Permutation and Combinationby ROHIT K - Saturday, 19 September 2009, 02:17 PM

Hi priya,

Color is the distinguishing factor and not the number... If we remove the color aspect from the argument only then the distinguishing property vanishes...

I do not complete ly agree with your conclusion..

The dice had two different colors viz. red and blue..(I did not remove them)

But, still you could not distinguish from each other..!!(how could you if they are the same ones?)



The reason was not color but same numerical value..If you had changed the number to then you would have distinguished it from occurrence.

if you can check my explanation again after reading this hope it would be more clear..

1. If colors of dice are same(i.e indistinguishable dice) then occurrence of or or for that matter any pair of values and viceversa cannot be distinguished..(very obvious) (indistinguishable because of color)

2. If colors of dice are different then

a. can be distinguished from occurrence of (distinguishable because of color of the dice)

b. or for that matter any same value on both dice cannot be distinguished from vice versa value(which is actually the sameoccurrence as the previous..) (indistinguishable because of same value on the dice)

Hence, the conclusion can be drawn that the distinguishing factor in the dice of different colors is the color aspect, only when numbersappearing on the dice are different. If the number becomes same(indistinguishable) then you cannot distinguish the occurrence.

Hope this helps..

Rohit

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Re: Permutation and Combinationby Suresh S - Monday, 21 September 2009, 11:11 AM

Hi sir ..its really superb.... thanks for ur article.....

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Re: Permutation and Combinationby ROHIT K - Tuesday, 22 September 2009, 12:36 PM

Hi TG..!!

Thanks for such a nice article.

I had one question regarding:

In how many ways can 4 persons be seated out of 5 boys and 3 girls in 4 different seats?

While expla ining you said it is not the simple formula based P&C problem.

Though, I think that it could be solved in very simple terms without break ing down the problem into cases(the way you already suggestedas the concept).

Two aspects to this problem:

1. Selecting 4 persons from 8 persons(5 boys+3 girls) which can be done in 8C4 ways

2. Arranging the 4 persons on 4 different seats for each se lection which can be done in 4! ways

So total number of ways = 8C4 * 4! = 70*24 = 1680 (same as 8P4)

Please let me know if this is not the right way to approach the problem.I need your insight into this.

Thanks and Regards

Rohit

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Re: Permutation and Combinationby Rohan S - Tuesday, 29 September 2009, 11:49 AM

TG STANDS FOR TRUE GUIDE........

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Re: Permutation and Combinationby sharmishtha Gupta - Friday, 9 October 2009, 01:13 PM

Hi,

superb article.

i have a doubt though,

in the ques where u hv 10 college students , 5 boys and 5 girls and u have to make sure that no two girls sit together,



can we first make the boys sit in 5! ways

_B_B_B_B_B_

now, for the girls to sit , we have to choose 5 of the 6 places available so that no two girls sit together, this can be done in 6 ways and thegirls can then be seated in 6* 5! ways.

Therefore, the no of ways in which they can sit according to the given condition then comes out to be

6*5!*5!.

P lease correct me if i am wrong.

Thanx

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Re: Permutation and Combinationby AsHwIn Drmz - Saturday, 10 October 2009, 01:32 AM

hi Sharmista,

You are missing out some hig here. Say 1st we seat a ll the boys..in 5! ways.

Then we have 6 places where 5 girls have to be seated.Now, 1st girl can be seated in any of the 6 available places,2nd girl can be seatedin the remaining 5 avaia lble places ,3rd girl can be seated in remaining 4 available places....etc

ie 6 x 5 x 4 x 3 x 2 .

so total arrangements boils down to : 2 * 5! *(6x5x4x3x2)

Hope its clear now.

Ashwin

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Re: Permutation and Combinationby sharmishtha Gupta - Saturday, 10 October 2009, 10:03 PM

Thanx ashwin.

U might think me to be a dud but k indly expla in how u got the

2 in 2*5!*(6*5*4*3*2) .

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Re: Permutation and Combinationby AsHwIn Drmz - Saturday, 10 October 2009, 10:59 PM

hey..

dont think like dat..We are all here to learn from eachother no one is perfect.So never underestimate urse lf...Coming to the question, seethere will be 2 cases :1st case when u seat a ll the boys 1st in 5 places then arrange the girls .2nd case you can seat the girls 1st thenarrange the boys among them.

hence we take 2*

Hope its clear now.

Ashwin

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Re: Permutation and Combinationby Saravanar B - Wednesday, 21 October 2009, 01:24 PM

Hi TG Sir,

Thanks a lot for your article and I don't have any words to describe ur work sir...

I owe a lot... Great job sir... I have a great respect for you sir and long live TG family

Have a small doubt sir :

1. Four friends go to a city in which there are 10 hotels. In how many ways can they stay ?



Sol :

(approach 1) : The first friend can stay in 10 ways, the second friend can stay in 10 ways, similarly the third

and fourth..

so the ans is 10*10*10*10 = 10000

(approach 2) : The first hotel can accomodate 4 friends or 3 friends or 2 friends or 1 friend or 0(zero). So nu

of ways the 1st hotel can accomodate is 1+2+3+4 = 10 ( i have added 1,2,3,4 as it is either - "OR" case)

Same goes for 2nd hotel, 3rd hotel and 4th hotel...

so the ans is 10*10*10*10 = 10000

*******************************************************

Now have a look at question no 2

*******************************************************

2. In how many ways can u post 10 letters in 4 letterboxes ?

Sol :

( approach 1 ) The first letter can be posted in 4 ways, similarly the second,third....tenth.

So the ans is : 4*4*4...*4 = 4^10 = 1048576

(approach 2 ) : The first letter box can get '1' or '2' or '3' ..... '10' letters.So nu of ways

the 1st letter can accomodate is 1+2+3+4+..+10 = 55

Same goes for 2nd letter 3rd letter and 4th letter

so the ans should be 55*55*55*55= 9150625

Doubt : I couldn't figure out what is wrong with my second approach.. Kindly explain me sir.. Also tell me

given a question when to apply which approach...

My prob is I never get satisfied with one approach sir.. I always try out different approach..

I am sure ur explanation would be helpful to many of the students like me...

Thanks

SaranShow parent | Reply

Re: Permutation and Combinationby Total Gadha - Wednesday, 21 October 2009, 09:42 PM

Hi Saravanar,

If first letter box posts even 1 letter or 2 letters etc. the second letter box does not have 10 ways.

Total Gadha

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Re: Permutation and Combinationby skysk iers 44 - Monday, 26 October 2009, 02:55 AM

Hello,



m not sure if this is the right section to post permutation problem, but anyways please help me out with the following

In how many ways can 10 soldiers stand in 2 rows such that there are 5 soldiers in each row?

Is the answer 7257600 or 7257600/2 ?

Puneet

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Re: Permutation and Combinationby Nabanshu Bhattacharjee - Tuesday, 27 October 2009, 01:33 AM

Hi TG

Thanks again,

Wanted to share the probabilistic approach to dearrangement problem with the TGites.

http://mathforum.org/library/drmath/view/56592.html

A query: If you go through the proof in the link I have given, you can see that they are considering 4 out of 5 letters placed in the correctenvelope. How come there result matches with the one given in this thread? I mean i could not find where I am going wrong. Please help!!!

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Re: Permutation and Combinationby Saravanar B - Friday, 30 October 2009, 12:26 AM

Hi TG,

Thanks for your reply...

Yes i got it... the second letter box need not have ten ways...

Is there a way to solve this question in the letter box perspective (approach 2) ?? I tried to solve but couldn't get the ans as well as the

approach..

Saran

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Re: Permutation and Combinationby swetha chinthareddy - Monday, 2 November 2009, 03:35 PM

6 Balls of different colours are to be distributed among 2 boys. What is the probability that each boy gets same number of balls?

a) 5/16b) 15/64c) 1/2d) 3/8

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Re: Permutation and Combinationby nishchai nevrekar - Wednesday, 4 November 2009, 05:58 PM

Hi,I would like to point out the that the solution to the 5 BOY , 5 GIRL MOVIE movie problem with no girls sitting together is a little flawed...

It doesnt take in to consideration this arrangement

GBGBGBBGBG n many others lik this....so the soln to this has to be considered by keeping objects (or people for this problem) who are not supposed to be together as dynamicand others as static...here boys have to be static n girls dynamic... as shown below

_B_B_B_B_B_so girls only have six positions to go into and also they will never be together as we lim iting them by providing only one slot btw boys. Hence refer to them as dynamic as they can take 6 positions.so, 6P5 = 6x120while boys who can be together are static cause they can only take 5 positions.5P5 = 120

so, total no ways = 120x120x6 = 14400x6

Hope this was helpful...Also, P lease correct me if I am wrong :D

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Re: Permutation and Combinationby Total Gadha - Friday, 6 November 2009, 01:46 PM



Hi Nishchal,

You forgot the condition mentioned in the question- every boy wants to sit with a girl.

Total Gadha

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Re: Permutation and Combinationby saurabh prabhudessai - Friday, 6 November 2009, 03:42 PM

in the condition mentioned by nischai every boy has at least one gal on his side!!

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Re: Permutation and Combinationby nishchai nevrekar - Friday, 6 November 2009, 03:59 PM

in these types of arrangements GB GB BG BG BGHere every boy is sitting with one girl.... it the condition was"every boy wants to sit in-between 2 girls" then i would be a diff case altogether.If not then plz temme... y these cases are not to be considered...

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Re: Permutation and Combinationby vineet ja in - Tuesday, 17 November 2009, 12:41 AM

Hi TG!!superb articleI have a question which has been asked by Rohit a lso.i am using same language as him

"In how many ways can 4 persons be seated out of 5 boys and 3 girls in 4 different seats?

While expla ining you said it is not the simple formula based P&C problem.

Though, I think that it could be solved in very simple terms without break ing down the problem into cases(the way you already suggestedas the concept).

Two aspects to this problem:

1. Selecting 4 persons from 8 persons(5 boys+3 girls) which can be done in 8C4 ways

2. Arranging the 4 persons on 4 different seats for each se lection which can be done in 4! ways

So total number of ways = 8C4 * 4! = 70*24 = 1680 (same as 8P4)"

please correct me if i am doing it wrong

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Re: Permutation and Combinationby SHAUNAK A - Tuesday, 24 November 2009, 08:16 PM

Hello TG sir,

That's CAT like article on Perm Com......simplicity exemplified...

Thanx Sir...

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Re: Permutation and Combinationby Harish A - Wednesday, 25 November 2009, 09:32 PM

Hi TG,

This is My first post in this is forum.

It is great disappointment I would like say, even after going through the Lessons multiple times, I am unable to solve even a singleproblem. It is not because of the way you expla in things, but it is my ability.

I am sure with these given set of sk ills, I am ill equipped to tack le CAT.

It may be of great help if you can let me where the solutions are present I can just read them for purpose of learning.

Thanks,Harish.A

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Re: Permutation and Combinationby aritreyee chaudhuri - Wednesday, 2 December 2009, 08:50 PM



Hi TG

I loved this article..thanks a lot for ur efforts ..this helped me recollect many of the concepts learnt earlier..

but i am not clear about the 'painting faces of a cube' problem..two other guys Syd S and Anirban Bhar have also asked the same questionthat i am going to type now..

When we fix the color of one side of a cube..we have 6 possibilities (as there are six colors)..and in that case the opposite side will have 5possibilities..for the remaining we can use the circular perrmutation rule..so the remaining sides can be painted in (4-1)! ways..

So shouldnt the number of ways in which the cube can be painted, be 6x5x3! ?

Can u please clarify this?

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Re: Permutation and Combinationby gurtez singh - Thursday, 11 February 2010, 05:56 PM

nice work sir ........... i hve a prob...

~ there r four sections of a paper wid a max. marks f 45 fr each section & to qualify 1 have get m in. 90 marks then in hw many ways he cnqualify da exam??????????????

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Re: Permutation and Combinationby vijayshree menon - Monday, 5 April 2010, 10:05 PM

Hi ,

Can any1 expla in me this

In the above post , TG sir cited an eg

@ an election meeting 100 ppl address rally. Order followed PM then MP followed by MLA

Solution given: Let PM = P MP denoted as Q and MLA as R. 10 speakers adddress ra lly in 10! ways. this includes PQR PRQ QPR QRP RPQRQP. Now Since need only PQR we divide 10! by 6.

I have a genuine prob here . First of a ll its decided tat PQR will be the order which means we need to keep PQR fixed and only count theremaining 7 ppl arrangement. Y cant it be so ??? I know I wrong where am I making the mistake ....

Im a real dumb one to have such doubts ... but want to make it clear ....

P ls TG sir or Dagny Mam or who ever cam amswer this q

Pls help Pls...

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Re: Permutation and Combinationby Aspirend Achieve - Monday, 10 May 2010, 02:27 PM

Hi TG Sir,

can you please expla in the solution for problem 3

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circular -linera queryby robin catch - Friday, 25 June 2010, 03:41 PM

DEAR TG..help me on this??

the number of ways that 4 girls and 5 boys can sit arounda table st no two girls sit together ?

is there a general formula??

what would be the solution for a linear table, ie 4 girls and5 boys( no 2 girls sit together)??

what would be the soln had 2 girls been always together,



(case1:around a round table

case 2: in a linear arrangement)???

thnx in advance, highly appreciate for quick reply

imlovinmath Show parent | Reply

Answersby Ramit Goyal - Monday, 28 June 2010, 11:04 PM

1 a2 b3 d4 c5 c6 a7 d8 c9 c10 b11 b12 c1314 a151617 b18 d19 a20 c

Please review and notify me for unattempted questions and wrong answers

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Re: Permutation and Combinationby sonali pawar - Thursday, 8 July 2010, 03:12 PM

Hi TG,

As usual beautiful article!!!!

Just one doubt......

In the part of Distribution in the problem of 5 balls and 3 boxes case IV i.e when all balls and boxes are dissim ilar, for distribution 113 and122 u say the no of distributions are 3!. Shouldn't it be 3!/2! = 3 as there are 2 groups with same number of things?

Please correct me if m wrong!

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Re: Permutation and Combinationby TG Team - Thursday, 8 July 2010, 05:32 PM

Hi Sonali

Though in 113 and 122 distribution there are two groups with same number of things but still they are differentgroups because of different balls. That's why it is 3! in place of 3!/2!

Hope it is c lear.

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Re: Permutation and Combinationby Naman Mirchandani - Tuesday, 20 July 2010, 04:29 PM

Sir,

I have one confusion:



In how many ways can 5 similar balls bedistributed in 3 different boxes?

Ans: (by method given here) 21

But there is one basic question in youarticle (like in how many can one post 10 lettersin 4 letter boxes: 4 * 4 *4.....10 times :4^10....).....

So if we apply the above method consideringsimilar letters and different letterboxes.....answer will come : 13C3 ?

Have we considered all letters dissimilar to get4^10? (can it be generalised where nothing ismentioned about similarity or dissimilarity ? )

Pls help ??????????? ....

NamanShow parent | Reply

Re: Permutation and Combinationby Siddharth Khanna - Saturday, 7 August 2010, 09:36 PM

Hi TG Sir ,

(

in these types of arrangements GB GB BG BG BGHere every boy is sitting with one girl.... it the condition was"every boy wants to sit in-between 2 girls" then i would be a diff case altogether.If not then plz temme... y these cases are not to be considered...)

Even I have the same doubt as mentioned by nishchai . Cant we consider this arrangement .

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Re: Permutation and Combinationby ravi baranwal - Sunday, 8 August 2010, 03:32 AM

agree wid siddharth n nischai.

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Re: Permutation and Combinationby kesav singh - Wednesday, 11 August 2010, 11:01 PM

hi everyone!



can anybody help me in below question?

how many different sums can be formed with the following coins

5 rupee,1 rupee,50 paise, 25 paise, 10 paise, 3 paisa , 2 paisa & 1 paise?

plz e leborate

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Re: Permutation and Combinationby Ravi Mathur - Tuesday, 17 August 2010, 09:16 PM

There are 10 steps in a sta ircase and a person has to take those steps. At every step the person has got a choice of tak ing 1 step or 2steps or 3 steps.The number of ways in whch person can take those steps is??

Can anyone please help me with this sta ircaseproblem.

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Re: Permutation and Combinationby versha jhunjhunwala - Sunday, 29 August 2010, 12:05 AM

Out of 21 tickets marked with numbers from 1 to 21 ,three are drawn at random, find the probability that the three numbers on them arein A.P

can u please expla in the complete method?

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Re: Permutation and Combinationby NITIN NIGAM - Tuesday, 31 August 2010, 07:57 AM

Hi Versha...my approach is as follows:

from 1 to 21 find the no. of AP's that u can get and there will be cases accordingly:

1. AP with d = 1 , we have 21 nos. Among these 21 nos. we can choose 3 nos. which are in AP as 1,2,3 or 2,3,4 or 3,4,5 and so on...hencetotal no. of cases will be 19 here with last case ending at 19,20,21

2. AP with d = 2, we have two series herea) 1,3,5,7......21 = 11 nos.hence nos. can be chosen as 1,3,5 or 3,5,7 and so on ..so here TOTAL CASES = 9b) 2,4,6,8......20 = 10 nos.hence nos. can be chosen as 2,4,6 or 4,6,8 and so on..so here TOTAL CASES = 8

3. AP with d = 3, we have the following seriesa) 1,4,7,10,13,16,19 -> total cases = 5b) 2,5,8,11,14,17,20 -> total cases = 5c) 3,6,9,12,15,18,21 -> total cases = 5

4. AP with d = 4, we have the following seriesa) 1,5,9,13,17,21 -> total cases = 4b) 2,6,10,14,18 -> total cases = 3c) 3,7,11,15,19 -> total cases = 3d) 4,8,12,16,20 -> total cases = 3

5. AP with d = 5, we have the following seriesa) 1,6,11,16,21 -> Total cases = 3b) 2,7,12,17 -> Total cases = 2c) 3,8,13,18 -> Total cases = 2d) 4,9,14,19 -> Total cases = 2e) 5,10,15,20 -> Total cases = 2

6. AP with d = 6, we have the following seriesa) 1,7,13,19 -> Total cases = 2b) 2,8,14,20 -> Total cases = 2c) 3,9,15,21 -> Total cases = 2d) 4,10,16 -> Total cases = 1e) 5,11,17 -> Total cases = 1f) 6,12,18 -> Total cases = 1

7. AP with d = 7, we have the following seriesa) 1,8,15 -> Total cases = 1b) 2,9,16 -> Total cases = 1c) 3,10,17 -> Total cases = 1d) 4,11,18 -> Total cases = 1e) 5,12,19 -> Total cases = 1f) 6,13,20 -> Total cases = 1g) 7,14,21 -> Total cases = 1

8. AP with d = 8, we have the following seriesa) 1,9,17 -> Total cases = 1b) 2,10,18 -> Total cases = 1c) 3,11,19 -> Total cases = 1d) 4,12,20 -> Total cases = 1e) 5,13,21 -> Total cases = 1

9. AP with d = 9, we have the following seriesa) 1,10,19 -> Total cases = 1b) 2,11,20 -> Total cases = 1



c) 3,12,21 -> Total cases = 1

10 AP with d = 10, we have the following seriesa) 1,11,21 -> Total cases = 1

now total no. of AP's cases that you can have from these nos. is19+17+15+13+11+9+7+5+3+1 = 100

now total no. of ways of se lecting 3 nos. from given 21 nos. are 21C3 = 1330

Hence the required probability is 100/1330

I could not think of a shorter approach but i certa inly fee l that there must be one....TGites please help here...k indly post a shorter orsimpler approach if you find one...and let me know if the answer is correct...

thanksNitin

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Re: Permutation and Combinationby deepti anand - Monday, 6 September 2010, 08:50 PM

hii...hv a ques..plz help me out

set A is formed by se lecting some of the numbers from the first 100 natural numbers such that the HCF of any two numbers in the set aresame

Q1:if every pair of set A has to be realtive ly prime and set A has max number of e lements possible,then in how many ways set A can beselected..??1)642)963)724)108

Q2: if the HCF of any two numbers in set A is 3, then what is the maximum no of e lements set A can have..??1)102)123)114)14

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Re: Permutation and Combinationby Animesh Devarshi - Saturday, 11 September 2010, 07:23 PM

Hi ,

reg question: how many ways can for persons be seated out of 5 boys and 3 girls on four different seats?

why can't we go for 8C4?

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Re: Permutation and Combinationby harleen mann - Monday, 4 October 2010, 01:27 AM

@deepti anand

Ans 1. think ing part: here, for e lements to be max, it can be understood that we need to take all primes numbers



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Re: Permutation and Combinationby Pritesh Ranjan - Tuesday, 19 October 2010, 03:11 PM

hi TG..i had 2 quests..i was wondering if u cud plz help me out wid dem..

1.)How many 6 digit numbers contain exactly 4 different digits..??

2.)Six white and Six black balls of the same size r 2 be distributed among 10 urns such that each urn contains at least one ball.What is thetotal no of distributions..????

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Re: Permutation and Combinationby Ankit Mitta l - Friday, 29 October 2010, 02:27 PM

Hi, I am ask ing this question to get my concepts right , here is the question :Determine the number of 5 card combinations out of a deck of 52 cards if thereis exactly one ace in each combination.

two ways of solving this: 4C1 X 48C4

&

4(ace can be se lected in 4 ways) X 48(next card can be se lected in 48 ways) X 47(next card can be se lected in 47 ways) X 46(..) X 45(..)

which one is the right way and why ?

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Re: Permutation and Combinationby TG Team - Friday, 29 October 2010, 04:00 PM

Hi Ankit

First one is correct.

In the selection of five cards their arrangement doesn't matter but in your second method the four cards have been

ordered and that needs to be divided by 4! which will again fetch the same result as calculated by first method.

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Re: Permutation and Combinationby Ankit Mitta l - Friday, 29 October 2010, 06:57 PM

Thanks very much for clearing the doubt , I got it right now for this one. Actually I am always confused when to consider the order and whennot.usually if arrangement is asked in the question I consider the order otherwise not.

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Re: Permutation and Combinationby saanthwana T - Thursday, 3 February 2011, 06:37 PM

This article really helped me to get a ll the basics of P & C

Thank you very much

It helped me get a ll the fundas.. so that I need not depend on 3 4 sources to get knowledge of single concept

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Re: Permutation and Combinationby saanthwana T - Tuesday, 8 March 2011, 09:30 PM

Hello,

I have small doubt

for permutations with sim ilar objects we use the formula

n!/ (k1! k2! k3! ....)

where n things are to be arranged, among which k1 is the number things of one typek2 is the number of things of 2nd typesoon ..........

now when among the n things above, if only r things are to be se lected and arranged then can we use the below formula?



nPr / (K1! K2!.....Ki!)

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Re: Permutation and Combinationby Neelkamal Biswas - Thursday, 28 April 2011, 10:57 AM

Hi TG,

In the solved example of 5 boys & 5 girls in a cinema, the two ways of sitting are given as : BGBGBGBGBG & GBGBGBGBGB

but we are not tak ing into account the possibility of GBGBGBGBBG as even in this situation no two girls sit together.The answer should then be : 5! * 6! as after arranging 5 boys there are six places in which the girls can be arranged _B_B_B_B_B_

Please advise on this .

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Re: Permutation and Combinationby TG Team - Thursday, 28 April 2011, 07:08 PM

Hi Neelkamal

In the given example: it should be the desire of the girls also that they won't let two boys sit together otherwise

your version is correct.

Kamal Lohia

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Re: Permutation and Combinationby deepti saluja - Monday, 2 May 2011, 12:16 PM

Hi Kamal,

Can u pls post the answers to the problems mentioned above (P n C), otherwise how do i check which 1 is right n which 1 is wrong. Pls help.

Thanks

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Re: Permutation and Combinationby TG Team - Monday, 2 May 2011, 04:40 PM

Hi Deepti

We intentionally don't provide the answers/solutions to problems so that you can always relish the taste of problemsolving. Also most of the problems have been discussed already many times in above posts. If you are unable to solvesome question or have some doubt in a particular question, you are more than welcome to discuss. But please don't

pressurize to provide answers to all the problems.

Kamal Lohia

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Re: Permutation and Combinationby deepti saluja - Tuesday, 3 May 2011, 05:53 PM

Hi Kamal,

Thanks for the quick reply.

Please help in these doubts:

Q4 license plates, i got 26^3 * 220 , what's the correct answer?

please make me understand, i could not solve these.

Q13 4 persons be chosen from a row of 10 persons,st no two persons are sitting next to each other.

Q15 Indian n Japanese teams

Q26 3 blocks of 5 seats each, 20 seats in a row..

Thanks N Regards,

Deepti

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Re: Permutation and Combination



by TG Team - Wednesday, 4 May 2011, 02:39 PM

Hi Deepti

4. Your answer is correct and surprisingly it's not given in the options. (Though we have already corrected in our assignments atTathaGat)

13. Lets make sit the remaining 6 persons of 10. Now we are to select 4 places only out of 7 available from where thechosen 4 persons were picked which can be easily done in C(7, 4) = 35 ways. Option (b) is correct.

15. We are to just determine the number of arrangements of I, I, I, I, J, J, J which is given by 7!/(4!)(3!) or C(7, 3) =35. Option (a) is correct.

26. Just place those 3 blocks of 5 seats each, so that you have used now 15 seats and remaining 5 seats are to bedistributed among the four gaps only. That's equivalent to find the number of whole number solution of a + b + c + d= 5 which is given by C(5 + 4 - 1, 4 - 1) = C(8, 3) = 56.

Hope this is c lear.

Kamal Lohia

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Re: Permutation and Combinationby nik ita dhanuka - Wednesday, 11 May 2011, 10:31 PM

Hi Kamal

i have one nagging doubt in circular arrangements..

suppose if we have 6 distinct beads. then 5!/2 distinct neck laces can be formed.how to go about a sim ilar sum if some of the beads are identical.the exact question is:

"A neck lace is to be made using 6 red, 3 blue and 3 green beads such that, no two red beads are adjacent to each other. All the beads ofthe same colour are identical. In how many ways can this neck lace be formed?"

the way i approached the sum...a ll the red beads can be arranged in only 1 way. this leaves 6 spaces between them. these spaces have tobe filled with the remaining 3 blue and 3 green beads. now i am stuck..this can be done in (5!/2*3!*3!)ways or (6!/2*3!*3!)ways??please expla in..

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Re: Permutation and Combinationby TG Team - Friday, 13 May 2011, 03:32 PM

Hi Nikita

In the numerator part there will be surely 5! and not 6! but why are you putting a 2 in denominator. Kindly expressyour thoughts.

Kamal Lohia

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Re: Permutation and Combinationby destiny unruled - Friday, 13 May 2011, 10:30 PM

Hi Nik ita

In case of circular permutations if we have identical things then it gets very complicated.

For the time being lets not consider the neck lace case and just consider the permutations of 6 red, 3 blue, and 3 green beads around acircle such that no two red beads are adjacent to each other.

Luck ily here we just have to arrange the blue and green beads between red beads as no two red beads can be together.

So, now the problem reduces to permuting 3 blue and 3 green beads around a circle.

Hence, answer should be 5!/(3!*3!), but its not an integer.

That means something is wrong.

So, lets go back to basics. Why do we divide by n while counting circular permutations to get (n - 1)! permutations.

Because we can get n different permutations by rotating a particular arrangement, but actually they are same. Thats why we divide by n.This is the case when we have all different objects(here we have "rotational period of n"*).

* - Rotational period of n means that if you rotate it n times it goes to an indistinguishable state from the first one

Now, suppose we have 3 red balls and 3 black balls. Now here we can have cases when we have rotational period of 1 or 2 or 3 or 6 (factorsof 6). But 1 is not possible coz then we should have all balls of same color.



Also 3 is not possible as then we need to have equal groups of three balls which means by Pigeonhole there must be at least 2 balls ofsome color Hence at least 4 balls of this color in a ll, which is not possible since there must be 3 of each color. Hence, only possibilities are 2 or 6

BGBGBG has a rotational period of 2, while a ll other cases have rotational period of 6.

Say there are x cases of rotational period of 6, then6*x + 2*1 = C(6, 3) = 20=> x = 3

=> There will be just 4 distinct permutations of 3 red balls and 3 black balls around a circle.

So, if you wants to solve a sim ilar k ind of question, then just find the number of different permutations of 6 black chairs and 6 white chairsaround a circle.

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Re: Permutation and Combinationby destiny unruled - Friday, 13 May 2011, 10:51 PM

Sorry, I forgot to solve it for neck lace.

In case of neck lace, BBGBGG and GGBGBB will a lso look alike, as now we can look at the neck lace from the back side also.

So, only 3 possible permutations.

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Re: Permutation and Combinationby tgdel156 tg - Sunday, 22 May 2011, 11:57 PM

Hello Sir !

P lease let me know the solution to questions involving the equation

like

1) 3x+ y + z= 40 where x,y,z are positive integer

2) x+y+z = 10 if x , y, z are distince positive integers.

3) x+y+z>= 10 ===> x+y+z -k =10

Thank You

Gaurav

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hi kamal

i have divided by 2 as the neck lace is inanimate and can be flipped over. so, for every 2 different circular arrangements, we actually haveonly one neck lace. hence, divided the total no. of possible circular arrangements by 2 to get the no. of distinct neck laces.even i was inclined towards 5!/3!*3!*2 as the answer. but, this is obviously wrong as it doesn't give an integer value. the book gives the answer as 6!/3!*3!*2. i am not being able to understand the logic behind this.i will appreciate if you can discuss your approach for this sum. and also if you can shed some light on these k ind of sums ingeneral...circular arrangement of things, some of which are identical.thanks,nik ita

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Hi Destiny

thanks for tak ing the time out. the book gives the answer as 6!/3!*3!*2 which works out to be 10.

i am still pretty confused in this regard.

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Re: Permutation and Combinationby destiny unruled - Thursday, 26 May 2011, 12:56 AM

Hi Nik ita

I'm pretty sure that the answer is 3 and not 10.

Did you get the logic I tried to expla in in my last post???

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Re: Permutation and Combinationby TG Team - Thursday, 26 May 2011, 05:36 PM

Hi Gaurav

1. x varies from 1 to 12 and number of solutions for x and y in each case form an AP. You just need to add them toget the total number of solutions. It comes out to be 3(1 + 2 + 3 + ... + 12) = 234.

2. You can easily count the ordered triplets by assigning values to x as 1, 2, ..., 7 - that comes out to be 24. OR just count the ordered triplets when all the three variables are positive integers i.e. C(9, 2) = 36. And subtractthe cases when any two of the variables are same (e.g. 11.., 22.., 33.., 44..). So removing 3*4 = 12 ordered tripltsfrom total 36, we are left with 24 ordered triplets of three distinct positive integers which add up to 10.

3. I am sure that there is some error in question because for the stated question answer is going to be infinitelylarge. See If x + y + z 10, that means the solutions when x + y + z = 10 will be added to the number of solutions of

x + y + z = 11,...and so on. So there is no end and the answer is (infinity) for sure.

Kamal Lohia

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Re: Permutation and Combinationby Nikunj Bajaj - Sunday, 21 August 2011, 02:56 PM

I have a doubt:-

In the question where we need to paint 6 faces of the cube with 6 different colors, why have we not multiplied the outcome with 6?

when we fixed the first color, it could have been done in 6 ways..

what m i m issing?

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Re: Permutation and Combinationby sowmya k - Monday, 22 August 2011, 11:39 PM

Hi TG

came across this forum damn too late but simply brilliant are your lessons and tips... as much about basics and foundation as they areabout tricks and shortcuts m inus any heavy duty formulae.. a lot of my friends too are hooked onto this site now enroute me ofcourse ;)

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Re: Permutation and Combinationby anupam chaturvedi - Wednesday, 24 August 2011, 03:07 AM

Hi TG,

Its a wonderful article.This is the first time I have been able to understand P&C, that fear has somewhat vanished now..!!

Still there are quiet a few things that I couldn't understand.

1) No. of ways of distributing 'n' sim ilar objects into 'r' sim ilar groups.

2) No. of ways of distributing 'n' distinct objects into 'r' sim ilar groups.

and a few questions that I'll post soon.

Looking ahead for your reply on above 2 problems!

Thanks a ton in adv. !

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Re: Permutation and Combinationby ketav sharma - Wednesday, 7 September 2011, 12:07 AM

come on man solved all questions where is the solution

I was highly disappointed after seeing no key

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Re: Permutation and Combinationby priyanka j - Wednesday, 7 September 2011, 11:50 AM

Sir,

1) Is d ans of "god boy bad boy movie" ques is 8*6! =5760



2) In d next ques of "election meeting" . How wud we know how many MP,MLA

pnc totalgadha

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