PERMUTATIONS AND

COMBINATIONS

BOTH

PERMUTATIONS AND COMBINATIONS

USE A COUNTING METHOD

CALLED FACTORIAL.

A FACTORIAL is a counting method that uses consecutive whole numbers as factors.

The factorial symbol is !

Examples 5! = 5x4x3x2x1

= 120

7! = 7x6x5x4x3x2x1

= 5040

First, we’ll do some permutation problems.

Permutations are “arrangements”.

Permutations

In the context of counting problems, arrangements are often called permutations; the number of permutations of n things taken r at a time is denoted nPr. Applying the fundamental counting principle to arrangements of this type gives

nPr = n(n – 1)(n – 2)…[n – (r – 1)].

Factorial Formula for Permutations

The number of permutations, or arrangements, of n distinct things taken r at a time, where r n, can be calculated as

!.

( )!n rn

Pn r

Example: Permutations

Evaluate each permutation.

a) 5P3 b) 6P6

5 35! 5!

a) 60(5 3)! 2!

P

6 66! 6!

b) 720(6 6)! 0!

P

Solution

Example: IDs

How many ways can you select two letters followed by three digits for an ID if repeats are not allowed?

Solution

26 2 10 3 650 720 468,000P P

There are two parts:1. Determine the set of two letters.2. Determine the set of three digits.

Part 1 Part 2

Example: Building Numbers From a Set of Digits

How many four-digit numbers can be written using the numbers from the set {1, 3, 5, 7, 9} if repetitions are not allowed?

5 45! 5!

120.(5 4)! 1!

P

SolutionRepetitions are not allowed and order is important, so we use permutations:

Let’s do a permutation problem.

How many different arrangements are there for 3 books on a shelf?

Books A,B, and C can be arranged in these ways:

ABC ACB BAC BCA CAB CBA

Six arrangements or 3! = 3x2x1 = 6

In a permutation, the order of the books is important.

Each different permutation is a different arrangement.

The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.

You try this one:

1. How many ways can 4 books be arranged on a shelf?

4! or 4x3x2x1 or 24 arrangements

Here are the 24 different arrangements:

ABCD ABDC ACBD ACDB ADBC ADCB

BACD BADC BCAD BCDA BDAC BDCA

CABD CADB CBAD CBDA CDAB CDBA

DABC DACB DBAC DBCA DCAB DCBA

Now we’re going to do 3 books on a shelf again, but this time we’re going

to choose them from a group of 8 books.

We’re going to have a lot more possibilities this time, because there

are many groups of 3 books to be chosen from the total 8, and there are

several different arrangements for each group of 3.

If we were looking for different arrangements for all 8 books,

then we would do 8!But we only want the

different arrangements for groups of 3 out of 8, so we’ll

do a partial factorial, 8x7x6=336

Try these:

1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?

10x9x8x7x6 or 30240

2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?

7x6x5x4 or 840

Combinations

Combinations

In the context of counting problems, subsets, where order of elements makes no difference, are often called combinations; the number of combinations of n things taken r at a time is denoted nCr.

Factorial Formula for Combinations

The number of combinations, or subsets, of n distinct things taken r at a time, where r n, can be calculated as

!.

! !( )!n r

n rP n

Cr r n r

Note: ! !

.!( )! ( )! !n r n n r

n nC C

r n r n r r

Example: Combinations

Evaluate each combination.

a) 5C3 b) 6C6

5 35! 5!

a) 103!(5 3)! 3!2!

C

6 66! 6!

b) 16!(6 6)! 6!0!

C

Solution

Now, we’ll do some combination problems.

Combinations are “selections”.

There are some problems where the order of the items is NOT important.

These are called combinations.

You are just making selections, not making different arrangements.

Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed?

Let’s call the 5 people A,B,C,D,and E.

Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.

Because we’re not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees.

Therefore instead of using only 5x4x3, to get the fewer committees, we must divide.

5x4x3

3x2x1

(Always divide by the factorial of the number of digits on top of the fraction.)

Answer:

10 committees

Now, you try.

1. How many possible committees of 2 people can be selected from a group of 8?

8x7

2x1or 28 possible committees

2. How many committees of 4 students could be formed from a group of 12 people?

12x11x10x9

4x3x2x1

or 495 possible committees

Example: Finding the Number of Subsets

Find the number of different subsets of size 3 in the set {m, a, t, h, r, o, c, k, s}.

SolutionA subset of size 3 must have 3 distinct elements, so repetitions are not allowed. Order is not important.

9 39! 9!

843!(9 3)! 3!6!

C

Example: Finding the Number of Poker Hands

A common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible?

SolutionRepetitions are not allowed and order is not important.

52 552! 52!

2,598,9605!(52 5)! 5!47!

C

Guidelines on Which Method to Use

Permutations Combinations

Number of ways of selecting r items out of n items

Repetitions are not allowed

Order is important. Order is not important.

Arrangements of n items taken r at a time

Subsets of n items taken r at a time

nPr = n!/(n – r)! nCr = n!/[ r!(n – r)!]

Clue words: arrangement, schedule, order

Clue words: group, sample, selection

Permutations and CombinationsEvaluate each problem.

c) 6P6a) 5P3b) 5C3 d) 6C6

54360 10 720 1

Permutations and CombinationsHow many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Two parts:

2. Determine the set of three digits.1. Determine the set of two letters.

26P2 10P3

2625650 1098720650720468,000

Permutations and CombinationsA common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible?

Hint: Repetitions are not allowed and order is not important.

52C5

2,598,960 5-card hands

Permutations and CombinationsFind the number of different subsets of size 3 in the set: {m, a, t, h, r, o, c, k, s}.

Find the number of arrangements of size 3 in the set: {m, a, t, h, r, o, c, k, s}.

9C3

84 Different subsets

9P3

987504 arrangements

10.3 – Using Permutations and CombinationsGuidelines on Which Method to Use

CIRCULAR PERMUTATIONS

When items are in a circular format, to find the number of different arrangements, divide:

n! / n

Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?

6! 6 = 120

Fundamental Counting Principle

For a college interview, Robert has to choose what to wear from the following: 4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he have to choose from?

4*3*2*5 = 120 outfits

A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

Permutations

A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

Practice:

2436028*29*30)!330(

!30330

27!

30! p

From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

Permutations

From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

Answer:

480,100,520*21*22*23*24

)!524(

!24524

19!

24! p

To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

CombinationsTo play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

Answer:

960,598,21*2*3*4*5

48*49*50*51*52

)!552(!5

!52552

5!47!

52! C

A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

CombinationsA student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

Answer:

101*2

4*5

)!35(!3

!535

3!2!

5! C

A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards?

CombinationsA basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards?

Answer:

2!1!1

!212 C

Center:10

1*2

4*5

!3!2

!525 C

Forwards:6

1*2

3*4

!2!2

!424 C

Guards:

Thus, the number of ways to select the starting line up is 2*10*6 = 120.

22512 * CCC 4*

How many ways can a student government select a president, vice president, secretary, and treasurer from a group of 6 people?This is the equivalent of selecting and arranging 4 items from 6.

= 6 • 5 • 4 • 3 = 360

Divide out common factors.

There are 360 ways to select the 4 people.

Substitute 6 for n and 4 for r in

How many ways can a stylist arrange 5 of 8 vases from left to right in a store display?

Divide out common factors.

= 8 • 7 • 6 • 5 • 4

= 6720

There are 6720 ways that the vases can be arranged.

Awards are given out at a costume party. How many ways can “most creative,” “silliest,” and “best” costume be awarded to 8 contestants if no one gets more than one award?

= 8 • 7 • 6

= 336

There are 336 ways to arrange the awards.

How many ways can a 2-digit number be formed by using only the digits 5–9 and by each digit being used only once?

= 5 • 4

= 20

There are 20 ways for the numbers to be formed.

1. Six different books will be displayed in the library window. How many different arrangements are there?

2. The code for a lock consists of 5 digits. The last number cannot be 0 or 1. How many different codes are possible? 80,000

720

3. The three best essays in a contest will receive gold, silver, and bronze stars. There are 10 essays. In how many ways can the prizes be awarded?

4. In a talent show, the top 3 performers of 15 will advance to the next round. In how many ways can this be done?

455

720

TournamentProblems

An amusement park has 27 different rides. If you have 21 ride tickets, how

many different combinations of rides can you take?

Answer:

296,010

Pop’s Pizza offers 4 types of meat and 3 types of cheese. In how many ways

could a pizza with two meats, different or double of the same meat, and one

cheese be ordered?

Answer:

30

Pop’s Pizza offers 4 types of meat and 3 types of cheese. In how many ways

could a pizza with two meats, different or double of the same meat, and one

cheese be ordered?

Answer:

30