Lesson 12.6, For use with pages 680-684

Evaluate.

1. 3! 2. 5!

Lesson 12.6, For use with pages 680-684

Evaluate.

ANSWER 6

1. 3! 2. 5!

ANSWER 120

Combinations

Section 12.6

Essential Questions

• What are the differences between permutations and combinations?

• What are the differences between odds and probability?

• How is probability used to make predictions?

• What are the differences between experimental and theoretical probabilities?

• In this section we will learn about selecting items when order is not important.

• Combination: is a group of items whose order is NOT important.

• For example – if I were to select three students from HR to serve on a committee. The order in which I selected these 3 does not matter. They form a committee or group. This is also called a combination.

• Students A, B, C & D (4) are trying out for a volleyball team. There are slots for only 3 players. What are the combinations of 3 players that could be chosen for the team?

• Make an organized list.

• A-B-C

• A-B-D

• A-C-D

• B-C-D

• There are only 4 combinations of 3 players chosen from a group of 4 players.

EXAMPLE 1 Listing Combinations

County Fair

You have 4 tickets to the county fair and can take 3 of your friends. You can choose from Abby (A), Brian (B), Chloe (C), and David (D). How many different choices of groups of friends do you have?

SOLUTION

List all possible arrangements of three friends. Then cross out any duplicate groupings that represent the same group of friends.

EXAMPLE 1 Listing Combinations

ABC, ACB, BAC,

BCA, CAB, and

CBA all represent

the same group.

ANSWER

You have 4 different choices of groups to take to the fair.

Combination

To find the number of combinations of n objects taken r at a time.

nCr = nPr

r!

If you selected 4 out of 10 books, the notation would look like this.

10C4 =10P4

4!

Combination

If you selected 4 out of 10 books, the notation would look like this.

10C4 10P4

4!= =

10 · 9 · 8 · 7

4 · 3 · 2 · 1= 210=

5040

24

EXAMPLE 2 Evaluating Combinations

Find the number of combinations.

Combination formulaa. 8C3 = 8P3

3!

= 56 Simplify.

3!= 8 · 7 · 6

= 8 · 7 · 6

3 · 2 · 1

Expand 3! = 3 · 2 · 1. Divide out common factors

(8 3)!8 P3=8!–

= 8 · 7 · 6

EXAMPLE 2 Evaluating Combinations

b. 9C 7 =9P7

7!Combination formula

9 P7 =9!

(9 7)!–

= 9·8·7·6·5·4·3

Expand 7!.

= 36 Simplify.

=9 · 8 · 7 · 6 · 5 · 4 · 3

7!

4

1

=9 · 8 · 7 · 6 · 5 · 4 · 3

7 · 6 · 5 · 4 · 3 · 2 · 1 Divide out common factors.

GUIDED PRACTICE for Example 2

Find the number of combinations.

2. 8C8 Combination formula

8 · 7 · 6 · 5 · 4 · 3 · 2 · 1=8!

8 C8 =(8 - 8)!

8! 8!0!=

= 1 Simplify.

8 · 7 · 6 · 5 · 4 · 3 · 2 · 1=8 · 7 · 6 · 5 · 4 · 3 · 2 · 1

Expand 8!.= 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1

8p8

8! =

Divide out common factors.

Tell whether the possibilities can be counted using a permutation or combination. Then write an expression for the number of possibilities.

3. You want to use a set of 8 lamps for a window display. Find how many sets you can choose from 25 lamps in the stock room.

ANSWER combination; 25C8 = 1,081,575

4. How many different ways can you select a preferred color and a substitute color from a mail-order catalog offering 12 colors of slacks?

ANSWER permutation; 12P2 = 132

Tell whether the possibilities can be counted using a permutation or combination. Then write an expression for the number of possibilities.

There are 8 swimmers in the 400 meter freestyle race. In how many ways can the swimmers finish first, second, and third?

Swimminga.

SOLUTION

Because the swimmers can finish first, second, or third, order is important. So the possibilities can be counted by evaluating 8P3 = 336

a.

EXAMPLE 3 Permutations and Combinations

Your track team has 6 runners available for the 4 person relay event. How many different 4 person teams can be chosen?

Trackb.

Order is not important in choosing the team members, so the possibilities can be counted by evaluating 6C4 = 15.

b.

SOLUTION

GUIDED PRACTICE for Example 3

A pizza shop offers 12 different pizza toppings. How many different 3 - topping pizzas are possible?

Pizza Toppings7.

Order is not important in choosing the team members, so the combination possibilities can be counted by evaluating 12C3 = 220.

SOLUTION

GUIDED PRACTICE for Example 3

Student Council

There are 15 members on the student council. In how many ways can they elect a president and a vice president for the council?

8.

SOLUTION

Because they elect a president and a vice president order is important. So the permutation possibilities can be counted by evaluating

15P2 = 210.

Homework

• Page 682 #1-21