Permutations, Combinations, and Counting Theory AII.12 The student will compute and distinguish between permutations and combinations and use technology.

Permutations, Combinations, and Counting Theory

AII.12 The student will compute and distinguish between permutations and combinations and use technology for applications.

Fundamental Counting Principle

If one decision can be made n ways and another can be made m ways, then the two decisions can be made nm ways.

In other words, to determine the number of ways independent events can happen together, find the product of the ways each event can happen.


A student is to flip a coin and roll a die. How many possible outcomes will there be?

Does flipping a coin effect the roll of a die or vice versa?

NO! Thus they are independent events.


A student is to flip a coin and roll a die. How many possible outcomes will there be?

How many outcomes when you flip a coin?

2 (heads or tails)How many outcomes when you

role a die?6 (1, 2, 3, 4, 5, or 6)


Thus the total number of outcomes is …

2 ● 6 = 12H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6


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?

There are four independent events (when you don’t consider fashion sense) : choice of slacks, choice of shirts, choice of shoes, choice of ties.


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?

Thus the total number of outfits (outcomes) is:

4 ● 3 ● 2 ● 5 = 120 possible outfits

Does Order Matter?

How four people finish a race

Picking people for a committee

Arranging letters to create words

Seating students in desks

Picking two of five desserts

Picking the starters for a game

Answering quiz questions

Permutations

An arrangement of items in a particular order.

Notice: ORDER MATTERS!!To find the number of permutations

of n items we can use the Fundamental Counting Principal or factorials.

Permutations – order matters

There are 8 acts in a talent show. Each act would prefer to be close to the start of the show. How many permutations are there for their order in the show?___ ___ ___ ___ ___ ___ ___ ___

The total number of ways the acts can be ordered is 40,320.

8 7 6● ● ● 5 ● 4 ● 3 ●

2 ● 1How many ways can you pick the 1st act?How many ways can you pick the 2nd act?How many ways can you pick the 3rd act?Continue this pattern to find the total number of ways to order the acts in the talent show.

FactorialTo find the number of outcomes

of 8 items in 8 positions, we multiplied 8 ● 7 ● 6 ● 5 ● 4 ● 3 ● 2 ● 1

This is 8! (8 factorial)To find a factorial, multiply the

given number by all the whole numbers less than it down to 1.

5! = 5 ● 4 ● 3 ● 2 ● 1 = 120Note 0! = 1


Twelve acts applied to be in the talent show, but time constraints only allow for eight acts. How many permutations of eight acts are possible in the talent show line up?

___ ___ ___ ___ ___ ___ ___ ___

= 19,958,400

How is this problem different from the last?

12 11 10● ● ● 9 ● 8 ● 7 ● 6 ● 5


___ ___ ___ ___ ___ ___ ___ ___The Fundamental Counting Principle is in use with permutations. Each place in the order is an independent event. Take the ‘ways’ each place can occur and multiply them to find the total number of outcomes. That is the Fundamental Counting Principle.

12 11 10● ● ● 9 ● 8 ● 7 ● 6 ● 5

Permutations - Official Formula

To find the number of Permutations of n items chosen r at a time, you can use the formula:

In the first problem, there were 8 acts and 8 places:

! where 0

( )!

nP r nn r n r

8 8

8! 8! 8!8! 40,320

(8 8)! 0! 1P

Permutations - Official Formula

In the first problem, there were 12 acts and 8 places:

. 0 where nrrn

nrpn

)!(

!

12 8

12! 12!

(12 8)! 4!

12 11 10 9 8 7 6 5 4 3 2 1

4 3 2 112 11 10 9 8 7 6 5 19,958,400

P


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?

30 3

30! 30!30 • 29• 28 24,360

(30 3)! 27!P

Answer Now


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?

24 5

24! 24!

(24 5)! 19!

24 • 23• 22 • 21• 20 5,100,480

P

Answer Now


There are three students left in the spelling bee – Arnold, Beth, and Corrie. Prizes are awarded for first and second place. How many different ways can these prizes be won? Create a list of possibilities.


3 ● 2 ● 1 = 61st – Arnold, 2nd – Beth1st – Arnold, 2nd – Corrie1st – Beth, 2nd – Arnold1st – Beth, 2nd – Carrie1st – Corrie, 2nd – Arnold1st – Corrie, 2nd – Beth

CombinationsArnold, Beth, and Corrie, all work

in the same department. One person must work on the 4th of July. How many different ways can the two people who get the day off be chosen?

3 – Arnold and Beth, Arnold and Corrie, Beth and Corrie.

CombinationsHow is this situation similar or

different from the last?Similar – choosing two of three

peopleDifferent – order did not matter,

we were looking for a ‘group’ Combinations are arrangements

of items where order does not matter.

Combinations – order doesn’t matter

In the Prize problem we listed six ways to give prizes. But if you notice, Arnold and Beth were paired twice, Arnold and Corrie were paired twice, and Beth and Corrie were paired twice. If the order did not matter, we would not need these repetitions.

Combinations are a subset of a permutation – we just through out the repetitions since the order is irrelevant.

In the July 4th problem, we had only 3 arrangements. If we take the 6 permutations from the Prize problem, and divide by 2 (each pairing happened twice) we get the 3 combinations.


To find the number of Combinations of n items chosen r at a time, you can use the formula:

Notice, this is the permutation formula divided by r! to cancel out the repetitions.

!0

!( )!

nC where r nn r r n r


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

Answer Now

5 3

5! 5! 5• 410

3!(5 3)! 3!2! 2 •1C


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?

Answer Now

52 5

52! 52!

5!(52 5)! 5!47!

52 •51•50• 49• 482,598,960

5• 4 •3• 2 •1

C


Calculator Calculations…

Casio – Run MenuOPTN keyMore options

(F6)PROB (F3)Permutation – F2Combination – F3Factorial – F1

TI – 84MATH keyArrow over to

PRBPermutation – 2Combination – 3Factorial - 4

Calculator Calculations: 7P4

Casio – Run MenuHit the OPTN keyMore options (F6)PROB (F3)Type the number 7Hit F2 for

permutationsType the number 4Hit EXE 840

TI – 84Type the number 7Hit the MATH keyArrow over to PRBEither arrow down

to #2 or type the number 2

Type the number 4Hit ENTER840

Practice …How many ways can 8 of 15

students line up to walk to lunch?How many ways can you choose

3 of 10 summer reading books?How many ways can pick a three

digit lock code if numbers cannot repeat?

How many ways can you pick 2 of 5 lamps for your new living room?

Practice …How many ways can 8 of 15

students line up to walk to lunch?Permutation; 15P8 = 259,459,200

How many ways can you choose 3 of 10 summer reading books?

Combination; 10C3 = 120

Practice …How many ways can pick a three

digit lock code if numbers cannot repeat?

Permutation; 10P3 = 720

How many ways can you pick 2 of 5 lamps for your new living room?

Combination; 5C2 = 10

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?

Answer Now

Challenge …

Challenge …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?

2 1

2!2

1!1!C

Center:

5 2

5! 5 410

2!3! 2 1C

Forwards:

4 2

4! 4 36

2!2! 2 1C

Guards:

Thus, the number of ways to select the starting line up is 120.

2 1 5 2 4 2 2 10 6 120 C C C

Challenge …A local restaurant is running a

dinner date special. For $24.99 you can choose 2 of 6 appetizers, 2 of 12 entrées, and 2 of 4 desserts. Assuming you and your date do not order the same items, how many ways could you create a dinner date special?

Challenge …A local restaurant is running a dinner date

special. For $24.99 you can choose 2 of 6 appetizers, 2 of 12 entrées, and 2 of 4 desserts. Assuming you and your date do not order the same items, how many ways could you create a dinner date special?

6 2

6!15

2!4!C

Appetizers:

12 2

12!66

2!10!C

Entrées:

4 2

4!6

2!2!C

Desserts:

Thus the number of possible meals is 5, 940.

6 2 12 2 4 2• • 15• 66 • 6 5,940C C C

Permutations, Combinations, and Counting Theory AII.12 The student will compute and distinguish between permutations and combinations and use technology.

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