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Math Coach Amy
Amy Ferguson Moncure

H Algebra 2/Trig · Unit 8

Combinations

Use a combination when the group matters and the order inside the group does not.

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Key Ideas Easy Medium Hard

nCr is a combination: order does not matter.

nCr = n!r! (n − r)!

nCr = nCn − r

What to Remember

Order does not matter

A combination chooses a group. Who is first inside the group does not create a new outcome.

Committees, teams, and identical prizes are combinations.

The smaller choose

nCr = nCn − r.

If r is more than half of n, switch to the smaller number before you multiply.

Level 1

Easy

One group

Example
A club of 9 members chooses a group of 2 to bring snacks. How many groups are possible?

nCr = n!r! (n − r)!

9C2 = 9 × 82 × 1

722 = 36

Answer: 36

1
A class of 8 students chooses a group of 3 for a poster. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

8C3 = 8 × 7 × 63 × 2 × 1

3366 = 56

Answer: 56

2
A shelf of 7 books will have 2 taken to the library desk. How many pairs are possible?
Show Answer

nCr = n!r! (n − r)!

7C2 = 7 × 62 × 1

422 = 21

Answer: 21

3
A team of 10 players chooses 3 to start a board. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

10C3 = 10 × 9 × 83 × 2 × 1

7206 = 120

Answer: 120

4
A cook has 6 spices and will use 2 of them. How many pairs are possible?
Show Answer

nCr = n!r! (n − r)!

6C2 = 6 × 52 × 1

302 = 15

Answer: 15

5
A choir of 11 students chooses 2 soloists who share one part. How many pairs are possible?
Show Answer

nCr = n!r! (n − r)!

11C2 = 11 × 102 × 1

1102 = 55

Answer: 55

Level 2

Medium

Choose more than half, then switch

Example
A group of 14 volunteers will send 11 of them to a workshop. How many groups are possible?

nCr = n!r! (n − r)!

14C11 = 14C3

14C3 = 14 × 13 × 123 × 2 × 1

2,1846 = 364

Answer: 364

1
A set of 15 photos will have 12 placed in one album. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

15C12 = 15C3

15C3 = 15 × 14 × 133 × 2 × 1

2,7306 = 455

Answer: 455

2
A list of 12 songs will have 9 kept on a playlist where order does not matter. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

12C9 = 12C3

12C3 = 12 × 11 × 103 × 2 × 1

1,3206 = 220

Answer: 220

3
A class of 16 students will choose 13 to attend one meeting. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

16C13 = 16C3

16C3 = 16 × 15 × 143 × 2 × 1

3,3606 = 560

Answer: 560

4
A stack of 17 cards will have 14 kept. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

17C14 = 17C3

17C3 = 17 × 16 × 153 × 2 × 1

4,0806 = 680

Answer: 680

5
A roster of 18 names will have 15 selected for one list. How many groups are possible?
Show Answer

nCr = n!r! (n − r)!

18C15 = 18C3

18C3 = 18 × 17 × 163 × 2 × 1

4,8966 = 816

Answer: 816

Level 3

Hard

Two groups at once

Example
A committee needs 2 coaches from 6 coaches and 3 players from 8 players. How many committees are possible?

Order does not matter in either group, so each group is a combination.

Ways = 6C2 × 8C3

nCr = n!r! (n − r)!

6C2 = 6 × 52 × 1

302 = 15

nCr = n!r! (n − r)!

8C3 = 8 × 7 × 63 × 2 × 1

3366 = 56

15 × 56 = 840

Answer: 840

1
A panel needs 2 teachers from 5 teachers and 3 students from 7 students. How many panels are possible?
Show Answer

Order does not matter in either group, so each group is a combination.

Ways = 5C2 × 7C3

nCr = n!r! (n − r)!

5C2 = 5 × 42 × 1

202 = 10

nCr = n!r! (n − r)!

7C3 = 7 × 6 × 53 × 2 × 1

2106 = 35

10 × 35 = 350

Answer: 350

2
A booth needs 3 painters from 4 painters and 2 builders from 9 builders. How many crews are possible?
Show Answer

Order does not matter in either group, so each group is a combination.

Ways = 4C3 × 9C2

nCr = n!r! (n − r)!

4C3 = 4C1

4C1 = 41

4

nCr = n!r! (n − r)!

9C2 = 9 × 82 × 1

722 = 36

4 × 36 = 144

Answer: 144

3
A trip needs 4 seniors from 7 seniors and 2 juniors from 6 juniors. How many groups are possible?
Show Answer

Order does not matter in either group, so each group is a combination.

Ways = 7C4 × 6C2

nCr = n!r! (n − r)!

7C4 = 7C3

7C3 = 7 × 6 × 53 × 2 × 1

2106 = 35

nCr = n!r! (n − r)!

6C2 = 6 × 52 × 1

302 = 15

35 × 15 = 525

Answer: 525

4
A lab group needs 2 mentors from 8 mentors and 3 students from 5 students. How many groups are possible?
Show Answer

Order does not matter in either group, so each group is a combination.

Ways = 8C2 × 5C3

nCr = n!r! (n − r)!

8C2 = 8 × 72 × 1

562 = 28

nCr = n!r! (n − r)!

5C3 = 5C2

5C2 = 5 × 42 × 1

202 = 10

28 × 10 = 280

Answer: 280

5
A display needs 4 posters from 9 posters and 2 banners from 4 banners. How many displays are possible?
Show Answer

Order does not matter in either group, so each group is a combination.

Ways = 9C4 × 4C2

nCr = n!r! (n − r)!

9C4 = 9 × 8 × 7 × 64 × 3 × 2 × 1

3,02424 = 126

nCr = n!r! (n − r)!

4C2 = 4 × 32 × 1

122 = 6

126 × 6 = 756

Answer: 756

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