In a total of 9 cases, a majority of case decisions against the lower court should be at least 5, 6, 7, 8, or 9. It can't be 4 or less as it would discourage the condition of the problem.

So, total no. of ways = 9C5 + 9C6 + 9C7 + 9C8 + 9C9

= 126 + 84 + 36 + 9 + 1

=256.

2)

Four consonants can be taken out from 12 in 12C4 ways.

And, three vowels can be taken out from 5 in 5C3 ways.

And, 4 consonants and 3 vowels are 7 letters. These can be arranged in 7! ways to form 7! words with or without meaning.

So, the number of different words that can be formed with 12 consonants and 5 vowels, by taking 4 consonants and 3 vowels in each word = 12C4 x 5C3 x 7!

............... = 495 x 10 x 7!

............... = 4950 x 7!.SHORT TRICk:-majority decisions occur when:= 5for + 6for + 7for + 8for + 9 for= 9C5 + 9C6 + 9C7 + 9C8 + 9C9= 126 + 84 + 36 + 9 + 1=256

2]number of different words= groups of consonants *groups of vowels*order of the letters= 12C4*5C3 *(4+3)!= 495*10*7!

The letters of the words CALCUTTA and AMERICA are arranged in all possible ways.?Calcutta has 8 letters, but there are 2 "C"s, 2"A"s and 2"T"s.If there were no duplicated letters, then there would be 8! ie 8x7x6x5x4x3x2x1 ways to arrange the letters.The first could be any of 8The second could be any of 7 (the first is already in place)The third could be any of 6 (the 1st and 2nd are already in place)and so on = 8x7x6x5x4x3x2x1 =40320 waysHowever we could swop each position where one "C" is for the other "C" and vice versa. Therefore the number of combinations is double what it should be in respect of having two "C"s.Similarly we need to divide by 2 for the "A"s and another divide by 2 for the 2 "T"s.Therefore Calcutta has 8! / 2*2*2 = 8x7x6x5x4x3x2x1 / 8 = 7!There are 5040 combinations for the letters of Calcutta

In a similar way America has 7 letters and 2 are "A"s.Therefore the number of combinations for America = 7! / 2 = 2520 combinations

Comparing the two: Calcutta has 7! combinations and America has 7!/2 combinationsThe ratio is therefore 5040 : 2520 or put another way 7! : 7!/2Multiply both sides by 2 and divide both sides by 7!

Calcutta : America = 2 : 1Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66B. 70C. 136D. 330E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.{abcd}- all 4 letters are distinct:C48=70;{aabc}- two letters are alike and other two are distinct:C13C27=63(C13is a # of ways to choose which two letters will be alike from 3 pairs andC27# of ways to choose other two distinct letters from 7 letters which are left);{aabb}- two letters are alike and other two letters are also alike:C23=3(C33is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Q.How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?

A.16C77!

B.12C44C37!

C.12C34C4

D.12C44C3

Solution:4 consonants out of 12 can be selected in12C4ways.

3 vowels can be selected in4C3ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels=12C44C3

Each group contains 7 letters, which can be arranging in7!ways.

Therefore required number of words=12C44C37!

8 guests have to be seated on a rectangular table, (4 on each side). 2 particular guests desire to sit on one side of the table and 3 on other side. The number of ways in which such sitting arrangement can be made is ? We have 2 choices available on one side and 1 choices available on other side. Let us start with one side. The number of choices are 2! * 3p2 = 2*6 = 12, However, these 2 groups (2 each) can exchange seats among themselves, thus we have 12*2 = 24. Let us now come to the other side of the table, we have 3! * 3C1 = 18 options. However, the 4 th person can take any seat out of 4 seats, so we have 18*4 = 72 Thus 72 *24 = 1728 answer.

There are 6 questions (with one choice) in a paper. In how many ways can I person attempt one or more questions? If I try to attempt all 6 questions, the choices are : 2*2*2*2*2*2 = 64 If I try to attempt any 5, then I have 5 choices 6c5, I can further take any choice out of these: 6C5*2*2*2*2*2 = 192 Similarly, if I try to attempt 4 out of these, I have 6C4*2*2*2*2 = 240 Similarly, if I try to solve 3 questions, I have 6C3*2*2*2 = 160 Similarly, if I try to attempt 2 questions, I have 6C2*2*2 = 60 Similarly, if I try to attempt only one question, 6C1*2 = 12 Let us add all these : 64+192+240+160+60+12 =728 answer.

Number of Letters/Characters in the word "COMMERCE"=8 {C, O, M, M, E, R, C, E}

nL=8

No. of Letters : a First Kind No. of C's = a a = a a Second Kind No. of M's = 2 b = 2 a Third Kind No. of E's = 2 c = 2 which are all different = 2 {O, R} x = 2nL= a + b + c + xIn the experiment of testing for the number of words that can be formed using the letters of the word "COMMERCE"Total No. of Possible Choices=Number of words that can be formed using the 8 letters of the word "COMMERCE"

n=nL

a! b! c!

=8!

2! 2! 2!

=8 7 6 5 4 3 2!

2 1 2 1 2!

=8 7 6 5 3

=5,040

Toolbox: C(n,r)=n!r!(nr)!No of questions in part I=6No of questions in part II=6The different ways of doing the questions are

(6C36C4)+(6C46C3)+(6C26C5)+(6C56C2)6C3=6!3!3!654321321321206C4=6!4!2!654321432121156C2=6!2!4!654!24!156C5=6!5!65!5!6(2015)+(1520)+(156)+(615)300+300+90+90780The results of 8 matches (win, loss or draw) are to be predicted. The no of different forecasts containingThe number of ways in which exactly 6 forecasts are correct is the number of ways of choosing 6 objects from a set of 8 multiplied by the number of predictions of wrong results for the last 2.i.e 8c2*2*2 = 112

The total number of all proper factors of 75600 is?We see that: 75600 = 2 * 3 * 5 * 7

Now to find the total number of factors, simply figure out all the different (distinct) ways you can choose any or all of those factors:

You have 4 distinct factors. You can choose 0 or more of each (and obviously only up to 4 2's, 3 3's, and 2 5's, and 1 7).

Think like this: Let's start with "choosing" 2. There are FIVE numbers that 2 can make (either 0, 1, 2, 3, or 4 2's: 1, 2, 4, 8, or 16). Now multiply that by however many numbers you can make from the rest. So now we have three 3's (so 4), which you can combine with ANY of the rest....do you see the pattern...the answer should be:

5 * 4 * 3 * 2 = 120 factors

How many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ?how many even numbers greater than 300 can be formed with digits 1,2,3,4,5 no digit being repeated ?

Ans:111

Not a GMAT question.

Number can be 3, 4, or 5 digit (4 and 5 digit number will obviously be more than 300).

5 digit even numbers:

Last digit must be either 2 or 4 (for number to be even), so 2 choices. First 4 digits can be arranged in 4! # of ways. So total # of 5 digit even numbers possible is4!2=48;

4 digit even numbers:

Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 3 digits out of 4 digits left when order of the digits matters isP34. So total # of 4 digit even numbers possible isP342=48;

3 digit even numbers:

Last digit again must be either 2 or 4, so 2 choices. # of ways to choose rest 2 digits out of 4 digits left when order of the digits matters isP24. So total # of 3 digit even numbers possible isP242=24.

But out of these 24 3-digit even numbers some will be less than 300, the ones starting with 1 or 2. There are 9 such numbers (1X2 and 3 options for the second digit X, so 3plus1X4 and 3 options for X, so 3plus2X4 and 3 options for X, so 3), which means that # of 3 digit even numbers more than 300 is249=15;

So total48+48+15=111.