PERMUTATIONS

COMBINATIONSAND

COUNTING TECHNIQUES

Computer Science, Statistics and Probability all involve counting techniques which are a branch of mathematics called combinatorics (ways to combine things). We'll be introducing this topic in this section.

For dinner you have the following choices:

soup salad chicken hamburgerprawns

icecream

ENTREES MAINS

DESSERTSHow many different combinations of meals could you make?

We'll build a tree diagram to show all of the choices.

soup

salad

chicken

prawnshamburger

chicken

hamburger

prawns

ice cream

ice cream

ice cream

ice cream

ice cream

ice cream

Now to get all possible choices we follow each path.

soup, chicken, ice cream

soup, chicken,

soup, prawns, ice cream

soup, prawns,

soup, hamburger, ice cream

soup, hamburger,

salad, chicken, ice creamsalad, chicken,

salad, prawns, ice cream

salad, prawns,

salad, hamburger, ice cream

salad, hamburger,

Notice the number of choices at each branch

2choices

3choices

2choices

We ended up with 12 possibilities

2 x 3 x 2 = 12

Multiplication Principle of Counting

If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in

different ways.

p x q x r

If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear?

6 x 4 x 5 x 3 = 360

Much quicker than making a list!

Choosing a path……how many ways from A to C through B?

A to B = 4 waysB to C = 2 ways4x2 = 8 ways

You try!

5 X 4 X 3 = 60 ways

A little tougher…..

A permutation is an ordered arrangement of r objects chosen from n objects.

For combinations order does not matter but for permutations it does.

There are three types of permutations.The first is distinct with repetition. This means there are n

distinct objects but in choosing r of them you can repeat an object.

this means different

Let's look at a 3 combination lock with numbers 0 through 9

There are 10 choices for the first number

There are 10 choices for the second number and you can repeat the first number

There are 10 choices for the third number and you can repeat

By the multiplication principle there are 10 x 10 x 10 = 1000 choices

Permutations: Distinct Objects with Repetition

The number of ordered arrangements of r objects chosen from n objects, in which the n objects are distinct and repetition is allowed, is nr

This can be generalized as:

What if the lock had four choices for numbers instead of three?

104 = 10 000 choices

The second type of permutation is distinct, without repetition.

Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition).

First place would be choosing someone from among 4 people.

Now there are only 3 to choose from for second place.

1st2nd

Now there are only 2 to choose from for third place. 3rd4th

Only one possibility for fourth place.

Based on the multiplication principle: 4 x 3 x 2 x 1 = 24 choices

nPr , means the number of ordered arrangements of r objects chosen from n distinct objects and repetition is not allowed.

In the last example:

0! = 1

If you have 10 people racing and only 1st, 2nd and 3rd place how many possible outcomes are there?

!!

rn

nPr

n

24!0

1234

!44

!44

4

P

720!7

!78910

!310

!103

10

P

A combination is an arrangement of r objects chosen from n objects regardless of order.

nCr , means the number combinations of r objects chosen from n distinct objects and repetition is not allowed.

Order doesn't matter here so the combination 1, 2, 3 is not different than 3, 2, 1 because they both contain the same numbers.

r

n

rrn

nCr

n or!!

!

Note: Dividing out by the common “r” combinations. Hence, you will have fewer combinations than permutations!

You need 2 people on your committee and you have 5 to choose from. You can see that this is without repetition because you can only choose a person once, and order doesn’t matter. You need 2 committee members but it doesn't matter who is chosen first. How many combinations are there?

102!3

!345

!2!25

!52

5

C

The third type of permutation is involving n objects that are not distinct.

How many different combinations of letters in specific order (but not necessarily English words) can be formed using ALL the letters in the word REARRANGE?

The "words" we form will have 9 letters so we need 9 spots to place the letters. Notice some of the letters repeat. We need to use R 3 times, A 2 times, E 2 times and N and G once.

First we choose positions for the R's. There are 9 positions and we'll choose 3, order doesn't matter

9C3

That leaves 6 positions for 2 A's.

R R R

6C2

A A

That leaves 4 positions for 2 E's.

4C2

That leaves 2 positions for the N.

E E

2C1

That leaves 1 position for the G.

N G

1C1

representative

example

84 15 6 2 1 = 15 120 possible "words"

Not Examinable.. Just for Fun

This can be generalized into the following:

Permutations Involving n Objects That Are Not Distinct

1 2

1 2

1 2

The number of permutations of objects

of which are of one kind, are of a

second kind, . . ., and are of a th kind

is given by

!

! ! !

where

k

k

k

n

n n

n k

n

n n n

n n n n

PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

Permutation:

Order Matters

We could have even numbers with 4, 5 or 6 digits

PART B: 4, 5 or 6 EVEN digits beginning with a 5

PART C: 5 or 6 EVEN digits beginning with a 2

PART D: 5 or 6 EVEN digits beginning with a 1 or 3

This Gives 4 possibilities to work with:

PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

4 2 22 4+2 2

This gives a total of 240

3 3

Permutation:

Order Matters

2 24 3 12+

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

4 3 31 4+1 2

This gives a total of 180

3 3

Permutation:

Order Matters

1 34 3 12+

PART B: 4, 5 or 6 EVEN digits beginning with a 5

PART C: 5 or 6 EVEN digits beginning with a 2

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

21 4 2

This gives a total of 96

3

Permutation:

Order Matters

1 24 3 12

+

PART D: 5 or 6 EVEN digits beginning with a 1 or 3

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

32 4 2

This gives a total of 288

3

Permutation:

Order Matters

2 34 3 12

+

PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 = 240

A Challenging Example. Have a go.

How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number?

Permutation:

Order Matters

We could have even numbers with 4, 5 or 6 digits

PART B: 4, 5 or 6 EVEN digits beginning with a 5 = 180

PART C: 5 or 6 EVEN digits beginning with a 2 = 96

PART D: 5 or 6 EVEN digits beginning with a 1 or 3 =288

This Gives 4 possibilities to work with:

Number of possible even numbers greater than 4000 = 804

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au