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Applied 40 Ch 4: Counting Methods

Chapter 4: Counting Methods

1. Counting Principle2. Permutations and Factorial Notation3. Permutations When all Objects are Distinguishable4. Permutations when Objects are Identicle5. Exploring Combinations6. Combinations7. Solving Counting Problems

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1. Counting Principle

There are 3 different ways of getting from Winnipeg to Brandon and 2 different ways to get from Brandon to Regina.

How many different ways are there to get from Winnipeg to Regina?

There are 3 ways to get from Winnipeg to Brandon, and for each route from Winnipeg to Brandon there are 2 routes from Brandon to Regina, that means there are 3 × 2 = 6 ways to get from Winnipeg to Regina, via Brandon.

How many ways are there to get from Winnipeg to Regina and back again?

3 ways from Winnipeg to Brandon,2 ways from Brandon to Regina,2 ways back from Regina to Brandon3 ways back from Brandon to Winnipeg

so there are 3 × 2 × 2 × 3 = 36 routes from Winnipeg to Regina and back again via Brandon.

What about if you cant use the same road twice?

3 ways from Winnipeg to Brandon,2 ways from Brandon to Regina,now there is only 1 way back from Regina to Brandonand 2 ways back from Brandon to Winnipeg

so there are 3 × 2 × 1 × 2 = 12 routes from Winnipeg to Regina and back again via Brandon without using the same road twice.

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There are lots of strategies that we can use to solve this type of problem.

eg. an outcome table

Brandon Route 1 Brandon Route 2Winnipeg Route 1 W1 B1 W1 B2Winnipeg Route 2 W2 B1 W2 B2Winnipeg Route 3 W3 B1 W3 B2

From the table we can see that there are 6 routes from Winnipeg to Regina via Brandon.

eg. a tree diagram

Again we can see that there are 6 routes from Winnipeg to Regina via Brandon from the tree diagram.

Or, we can use the Fundamental Counting Principle.

The Fundamental Counting Principle states that

if there are n ways to perform one task, and m ways to perform another, then there are n × m ways of performing both tasks.

and this can be extended for any number of tasks.

Example: You have 3 choices of starters, 5 main courses and 4 desserts. How many different suppers are possible?

starter main dessert# of possible suppers = 3 × 5 × 4 = 60

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Example: Hannah plays soccer on her school team. Their team uniform has:

3 different sweaters (red, white and black)3 different shorts (red, white and black)

How many different variations of the soccer uniform are there?

sweater shortchoices choices 3 × 3 = 9 variations

Example: A luggage lock opens with the correct 3 digit code. Each wheel rotates through the digits 0 to 9.

a) How many different 3 digit-codes are possible?b) Suppose each digit can only be used once in each code. How many codes are

possible if no repetition is allowed?

Solution

a) 1st digit 2nd digit 3rd digit10 × 10 × 10 = 1000 possible 3-digit codes

b) 1st digit 2nd digit 3rd digit 10 × 9 × 8 = 720 possible 3-digit codes

It is not always possible to use the Fundamental Counting Principle.

Example: Given a standard pack of playing cards.

How many ways are there of choosing:-a) Either a black card or an aceb) Either a red card or a 10

Solution

a) There are 26 black cards including the ace of clubs and the ace of spadesThere are 4 aces, including the ace of clubs and the ace of spades

You can not count the 2 aces twice so there are 28 ways that a black card or an ace can be chosen.

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b) There are 26 red cards including the ten of diamonds and the ten of heartsThere are 4 tens, including the ten of diamonds and the ten of hearts

You can not count the 2 tens twice so there are 28 ways that a red card or a ten can be chosen.

We have not used the Fundamental counting principle!

Summary

The Fundamental Counting Principle can be used when tasks are related by the word AND.

It can not be used for tasks related by the word OR.

For disjoint (mutually exclusive) sets, A and Bn ( A∪B )=n ( A )+n (B)

For non-disjoint sets C and D, using the principle of inclusion and exclusion

n (C∪D )=n (C )+n (D )−n(C∩D)

The Fundamental Counting Principle States:

If there are n ways to perform one task and m ways to perform another task then there are

n × m ways of performing both tasks.This can be extended for any number of tasks.

Organised lists, outcome tables and tree diagrams can be useful when solving counting problems as they list all the possible outcomes but can be difficult to use in tasks that have many outcomes.

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2. Permutations and Factorial Notation

How many different single file line-ups are possible when Melissa, Nikki and Frankie line up at the cashier of a fast food chain.

You could use a list:

Melissa, Nikki, FrankieMelissa, Frankie, NikkiFrankie, Melissa, Nikki 6 possible lines upsFrankie, Nikki, MelissaNikki ,Frankie, MelissaNikki, Melissa, Frankie

or use the Fundamental Counting Principle

# of possible # of possible # of possiblepeople in people in people in 1st place 2nd place 3rd place

3 × 2 × 1 = 6 possible line ups

This is actually an example of a permutation.

A permutation is an arrangement of distinguishable objects in a definite order.

Example: Determine the number of arrangements that 6 children can form while lining up for a drink.

SolutionThere are 6 possible positions.

Total number of lines ups = 6 × 5 × 4 × 3 × 2 × 1 = 720

A more convenient way of writing this is using factorial notations. 6! (6 factorial) = 6 × 5 × 4 × 3 × 2 × 1

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Factorial Notation- a concise way of showing the product of consecutive descending natural

numbers.

5! = 5 × 4 × 3 × 2 × 1 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

or n! = n × (n-1) × (n-2) × ( n-3) … × 3 × 2 × 1

NB: -2! and ½ ! have no meaningbut 0! = 1

Example: Evaluatea) 10!

b)12!

(9 !3 !)

Solution :

a) Using the ! button on your calculator

10! =

b)12!

(9 !3 !)=12×11×10×9 !

9 !×3×2×1=2×11×10=220

or use your calculator!

Example:

Simplify

a) 5 x 4! 5 x 4! = 5 x 4 x 3 x 2 x 1 = 5!

So using this as a pattern we can see that (n+3)(n+2)! = (n+3)!

b)9 !7 !

=9×8×7 !7 !

=9×8=72

c)So using this pattern

(n+1 ) !(n−1 ) !

= (n+1 ) (n ) (n−1 )!(n−1 ) !

= (n+1 )n=n2+n

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Example:

Solve

n!(n−2 )!

=90

Solutionn!

(n−2 )!=90

n (n−1 ) (n−2 )!(n−2)!

=90

n(n-1) = 90

n and (n-1) are two consecutive numbers so since 10 x 9 = 90, n = 10

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3. Permutations When all Objects are Distinguishable

How many 3-letter permutations can you make with the letters in the word MATH?

We could do this by listing them out, or using the fundamental counting principle

There are three possible positions

i.e. 4 x 3 x 2 = 24 possible arrangements

We could write this as

This leads us to a formula for the number of possible arrangements when r distinct objects are chosen from n distinct objects

where n is total number of objectsand r is number of objects being chosen

ExampleMatt downloaded 10 new songs from an online music store and wanted to make a

playlist using 6 of the songs in any order. How many ways could he do this?

SolutionThere are 10 songs to choose from for the 1st songthere are 9 songs to choose from for the 2nd songthere are 8 songs to choose from for the 3rd songthere are 7 songs to choose from for the 4th songthere are 6 songs to choose from for the 5th songthere are 5 songs to choose from for the 6th song

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i.e. 10 x 9 x 8 x 7 x 6 x 5 = 151 200

or using nPr = n !

(n−r )!= 10 !

(10−6 ) !=10 !

4 !=151200

or using the button on your calculator 10P6 = 151 200

ExampleAt a used car lot seven different car models are to be parked close to the street for easy viewing.How many ways can the cars be parked if:a) The three red cars must be parked so that there is a red car at each end and the third red car exactly in the middle. b) The three red cars must be parked side by side.

Solution:a) The order must be

Red-Car Car Car Red-Car Car Car Red-Car

There are 3 red cars available for 1st positionThere are 4 cars that aren’t red available for 2nd positionThere are 3 cars that aren’t red available for 3rd positionThere are now 2 red cars available for 4th position There are 2 cars that aren’t red available for 5th positionThere are 1 car, that isn’t red, left for 6th positionThere are 1 red car left for 7th positioni.e.

Red-Car Car Car Red-Car Car Car Red-Car 3 x 4 x 3 x 2 x 2 x 1 x 1

b) The three red cars must be parked together and can be treated as 1 unit. There are 3! ways of arranging the red cars within the unit.

That means we have 5 units to arrange, the four non-red cars and the 3-red-car unit. There are 5! ways if arranging these 5 units.

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Thus there are 5! x 3! = 720 ways of arranging the cars in this way.Example

A Canadian social insurance number consists of a 9-digit number that used digits from 0 to 9. How many SIN numbers can be created a) if there are no restrictions on the digits selected (i.e. repetition of digits is allowed)b) if digits can only be used once. (no repetition is allowed)

Solutiona) There are 10 possible digits for each of the 9 places in the SIN numbers so

10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 109

b) If the digits can only be used once thenthere are 10 digits available for the 1st place

9 digits available for 2nd place 8 digits available for 3rd place 7 digits available for 4th place 6 digits available for 5th place 5 digits available for 6th place 4 digits available for 7th place 3 digits available for 8th place 2 digits available for 9th place

i.e. 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3 628 800

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4. Permutations when Objects are Identical

How many ways can the word BIKE be arranged.

BIKE BIEK BEIK BEKI BKEI BKIEIBKE IBEK IKBE IKEB IEKB IEBKKIBE KIEB KEIB KEBI KBEI KBIEEIKB EIBK EBIK EBKI EKBI EKIB

or 4! = 4 x 3 x 2 x 1 = 24

How many ways can the word BOOK be arrangedBOOK BOKO BKOOOOBK OOKB OKOB OKBO OBOK OBKOKOOB KBOO KOBO

i.e. because the letter O is being repeated the number of arrangements has decreased and there are 12 arrangements of the 4 letters

ExampleHow many ways are there to arrange the letters in the words:

VACUUM, HANNAH, MISSISSIPPI

Solution

VACUUM 6 letter, U is repeated twice

# of arrangements = 6 !2!

=7202

=360

HANNAH 6 letters, H is repeated twice, A is repeated twice, N is repeated twice

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# of arrangements = 6 !

2!2 !2 != 720

2×2×2=90

MISSISSIPPI 11 letter, I is repeated 4 times, S is repeated 4 times, P is repeated twice

# of arrangements = 11!

4 ! 4 !2 !=34 650

ExampleHow many ways can 3 white, 1 red, 1 blue, 1 yellow and 1 green plates be

stacked?

Solution:There are 6 plates in total so if they were all different there would be 6! There are

3 white plates so we need to divide by 3!

# of arrangements = 7 !3!

=840

What about if there were 3 white, 2 red, 4 blue, 1 yellow and 1 green plates?

# of arrangements = 11!

3! 2! 4 !=138600

ExampleIt is common to see prayer flags in the mountainous regions of China, Nepal, India and Bhutan. Each flag has a prayer written on it and colour is used to represent different elements, green (water), yellow (earth), white (air/wind), blue (sky/space), and red (fire).

How many different arrangements of the same prayer can Danni make using these 9 flags: 1 green, 1 yellow, 2 white, 3 blue and 2 red?

Solution

# of prayers (arrangements) = 9 !

2!3 !2 !=15120

ExampleHow many ways can the letters of the word CANADA be arranged if the first letter must be N and the last letter must be C.

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SolutionDeal with the restrictions first.There is only one way we can put N first and one way to put a C last.

N _ _ _ _ C

That leaves 4 letters, 3As and a D so there are 4 !3 ! ways of arranging these remaining

letters

so # arrangements = 1×4 !3 !×1=4

ExampleJulie's home is 3 blocks north and 5 blocks west of her school. How many routes can Julie take from home to school if she always travels either South or East?

Solution We can solve this using the diagram by counting how many way there are to get to each “corner”

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or by considering this as an arrangement of 3 Souths and 5 Easts.

# of arrangements = 8 !

3!5 !=56

5. Exploring Combinations

The School athletic council has 5 members, Jill, Ted, Robin, Yvette and Martin. They are holding a bake sale next week.

How many ways can you select a committee of 2 from the council, to sell the baked goods?We could list the results

JT, JR, JY, JMTR, TY, TMRY, RMYM

For this problem it is important to realise that choosing Jill and Ted is the same as choosing Ted and Jill

This is an example of a combination. A combination is when the order of objects being grouped together doesn't matter.

ExampleAll the diamonds from a standard pack of cards are shuffled and 4 of these 13 cards are dealt out in a row. a) How many different permutations of 4 diamonds could be dealt.

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b) Suppose the order doesn't matter. How many different combinations could be dealt?

Solutiona) there are 13 diamond cards so

13 x 12 x 11 x 10 = 17 160 or

13P4 = 17 160

b) since there are 4! arrangements of the 4 cards we have to divide our answer in a) by this to get the number of arrangements when the order isn’t important.

i.e. 17 160

4 !=715

6. Combinations

The formula for finding the number of combinations is

where n is the total number of objects and r is the number of objects being chosen.

ExampleHow many ways can a team of three snow sculptors be chosen from 9 volunteers?

Solution The order doesn’t matter so this is a Combination

n = 9 (total of 9 volunteers)r = 3 (number being chosen)

# of possible teams = 9C3 = 9 !

3! (9−3 )!= 9 !

3 !6 !=84

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ExampleA restaurant serves 10 flavours of ice-cream. Danielle has ordered a large sundae with 3 scoops of ice cream. How many different ice-cream sundaes does Danielle have to choose from?

SolutionThe order doesn’t matter so this is a combination problemn = 10 (number of flavours available)r = 3 (# of flavours being chosen)# of possible sundaes = 10C3 = 120

Example Tanya is the coach of a Pole Push team that consists of 9 members, 5 male and 4 female. In each competition teams of 4 compete against each other to push their opponents out of the circle.

a) How many different 4 person teams are there for an all male competition?b) How many 4 person teams are there if there have two males and two females on the team.

Solutiona) There are 5 male players so n = 5 and we are choosing 4 team members so

# of all male teams = 5C4 = 5

b) There are 5 male players and we want 2 male team members so there are 5C2 ways of choosing the male members of the team.There are 4 female players and we want 2 female team members so there are 4C2 ways of choosing the male members of the team.# of teams consisting of 2 males and 2 females

= 5C2 × 4C2

= 10 × 6 = 60

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Example A planning committee is to be formed for a school wide Earth day program. There are 13 volunteers, 8 teachers and 5 students. How many ways can the head teacher choose the 4 person committee that has at least 1 teacher?

SolutionThis scenario has to be broken down into different cases because of the wording “at least” so a committee could consist of

Case 1: 3 students and 1 teacherCase 2: 2 students and 2 teachersCase 3: 1 student and 3 teachersCase 4: 0 students and 4 teachers.

Method 1Consider each case individually.

Case 1: 3 students and 1 teacherChoosing 3 students (r = 3) from 5 students (n=5) #of ways = 5C3

Choosing 1 teacher (r = 1) from 8 teachers (n=8) #of ways = 8C1

# of committees with 3 students and 1 teacher = 5C3 × 8C1 = 10 × 8 = 80 Case 2: 2 students and 2 teachers

Choosing 2 students (r = 2) from 5 students (n=5) #of ways = 5C2

Choosing 2 teachers (r = 2) from 8 teachers (n=8) #of ways = 8C2

# of committees with 2 students and 2 teachers = 5C2 × 8C2 = 10 × 28 = 280

Case 3: 1 student and 3 teachersChoosing 1 students (r = 1) from 5 students (n=5) #of ways = 5C1


# of committees with 1 student and 3 teachers = 5C1 × 8C3 = 5 × 56 = 280

Case 4: 0 student and 4 teachersChoosing 0 students (r = 3) from 5 students (n=5) #of ways = 5C0


# of committees with 0 students and 4 teachers = 5C0 × 8C4 = 1 × 70 = 70

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# of 4 person committees with at least one teacher = Case 1 + Case 2 + Case 3 + Case 4= 80 + 280 + 280 + 70= 710

Method 2# of 4 person committees with at least one teacher

= total # of committees - # of committees with 0 teachers

8 teachers and 5 students = 13 peopleTotal # of committees = 13C4 = 715

All student committees = 5C4 = 5

# of 4 person committees with at least one teacher = 715 – 5 = 710 committees

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7. Solving Counting Problems - using permutations and combinations

REMEMBERUse Permutations when the order is important, Use Combinations when the order is not important.

In each problem look for- If there is repetition you will need to divide- If multiple tasks/stages are linked by the word “and” use the Fundamental Counting Principle (i.e. multiply)- If multiple tasks/stages are linked by the word “or” the problem can usually be broken down into mutually exclusive events (add)

ExampleA piano teacher and her students are having a group photograph taken. There are 3 boys and 5 girls.

a) The photographer wants the boys to sit together and the girls to sit together for one of the poses. How many ways can the students and teacher sit in a row of nine chairs for this pose.

b) For a second pose the photographer wants the two tallest students Jill and Sam to sit at either end, Jill on the left and Sam on the right with the teacher in the middle. How many seating arrangements are there for this pose?

Solutiona) Treat the 3 boys as 1 unit.

There are 3! ways of arranging the boys within the unit.

Treat the 4 girls as 1 unit.There are 4! ways if arranging the girls within the unit

There is 1 teacher so there is 1! ways of arranging the teacher unit.

There are 3! ways of arranging the 3 units (Boys, Girls and Teacher)

Total # of arrangements = 3! × 3! × 4! × 1! = 4320

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b) Deal with the restrictions to begin with

There is only 1 way to seat Jill on the leftThere is only 1 way to seat Sam on the rightThere is only 1 way to seat the teacher in the middle seat

Jill __ __ __ Teacher __ __ __ Sam1 1 1

There are 6 students available to sit next to JillThere are 5 students available to sit in the next positionThere are 4 students available to sit in the next positionThere are 3 students available to sit in the position on right of the teacherThere are 2 students available to sit in the next positionThere are 1 student available to sit in the next position

Jill __ __ __ Teacher __ __ __ Sam1 × 6 × 5 × 4 × 1 × 3 × 2 × 1 × 1

= 6! = 720

Example Combination problems are common in computer science. Suppose there are 10 different data items (a,b,c,d,e,f,g,h,I,j) to be placed into 4 different memory cells in a computer.

Only 3 items are to be placed in the first cell, 4 data items in the second cell, 2 data items in the third data cell and 1 data item in the last cell.

How many ways can the ten data items be placed in the 4 memory cells

SolutionConsider each cell in turn.

Cell #1 10 data items available, 3 needed. # of combinations = 10C3 = 120

Cell #2 7 data items available (3 have been used), 4 needed. # of combinations = 7C4 = 35

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Total # of combinations (Cell 1 and Cell 2 and Cell 3 and Cell 4)= 120 × 35 × 3 × 1= 12 600

Example How many different five-card hands that contain at most one black card can be dealt to one person from a standard deck of playing cards.

SolutionAt most 1 black means 2 scenarios, a hand with 0 black cards or a hand with 1 black card

A standard deck contains 26 black cards and 26 red cards.

Case 1: 0 black cards (n = 26, r = 0) 5 red cards (n = 26, r = 5)26C0 × 26C5 = 1 × 65780 = 65 780

Case 2: 1 black cards (n = 26, r = 1), 4 red cards (n = 26, r = 4)26C1 × 26C4 = 26 × 14950 = 388 700

Total # hands = Case 1 or Case 2= 65 780 + 388 700 = 454 480

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