Here are some carefully chosen combination word problems that will show you how to solve word problems involving combinations.

Use the combination formula shown below when the order does not matter

The combination word problems will show you how to do the followings:

- Use the combination formula

- Use the multiplication principle and the combination formula

- Use the addition principle and the combination formula

- Use the multiplication principle, addition principle, and combination formula

Word problem #1

There are 18 students in a classroom. How many different eleven-person students can be chosen to play in a soccer team?

**Solution**

The order in which students are listed once the students are chosen does not distinguish one student from another. You need the number of combinations of 18 potential students chosen 11 at a time.

Evaluate _{n}C_{r} with n = 18 and r = 11

n!
r!(n - r)!

18!
11!(18 - 11)!

18!
11!(7)!

18×17×16×15×14×13×12×11!
11!(7×6×5×4×3×2)

18×17×16×15×14×13×12
(7×6×5×4×3×2)

160392960
5040

_{18}C_{11} = 31824

There are 31824 different eleven-person students that can be chosen from a group of 18 students.

**Word problem #2**

For your biology report, you can choose to write about three of a list of four different animals. Find the number of combinations possible for your report.

**Solution**

The order in which you write about these 3 animals does not matter as long as you write about 3 animals.

Evaluate _{n}C_{r} with n = 4 and r = 3

4!
3!(4 - 3)!

4!
3!(1)!

4×3×2
3×2(1)

24
6

= 4
There are 4 different ways you can choose 3 animals from a list of 4.

**Word problem #3**

A math teacher would like to test the usefulness of a new math game on 4 of the 10 students in the classroom. How many different ways can the teacher pick students?

**Solution**

The order in which the teacher picks students does not matter.

Evaluate _{n}C_{r} with n = 10 and r = 4

10!
4!(10 - 4)!

10!
4!(6)!

10×9×8×7×6!
4×3×2(6)!

10×9×8×7
4×3×2

5040
24

= 210
There are 210 ways the teacher can pick students

These combination word problems will also show you how to use the multiplication principle and the addition principle.

**Word problem #4**

A company has 20 male employees and 30 female employees. A grievance committee is to be established. If the committee will have 3 male employees and 2 female employees, how many ways can the committee be chosen?

**Solution**

This problem has the following two tasks:

**Task 1**: choose 3 males from 20 male employees

**Task 2:** choose 2 females from 30 female employees

We need to use the fundamental counting principle, also called the multiplication principle, since we have more than 1 task.

**Fundamental counting principle**

If you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks.

Therefore, evaluate _{20}C_{3} and _{30}C_{2} and then multiply _{20}C_{3} by _{30}C_{2}

20!
3!(20 - 3)!

20!
3!(17)!

20×19×18×17!
3×2(17)!

20×19×18
3×2

6840
6

= 1140
30!
2!(30 - 2)!

30!
2!(28)!

30×29×28!
2×1(28)!

30×29
2

870
2

= 435
_{20}C_{3} × _{30}C_{2} = 1140 × 435 = 495900

The number of ways the committee can be chosen is 495900

**Word problem #5**

Eight candidates are competing to get a job at a prestigious company. The company has the freedom to choose as many as two candidates. In how many ways can the company choose two or fewer candidates. **Solution**

The company can choose 2 people, 1 person, or none.

Notice that this time we need to use the addition principle as opposed to using the multiplication principle.

What is the difference? The key difference here is that the company will choose either 2, 1, or none. The company will not choose 2 people and 1 person at the same time. This does not make sense!

**Addition principle**

Let A and B be two events that cannot happen together. If n is the number of choices for A and m is the number of choices for B, then n + m is the number of choices for A and B.

Therefore you need to evaluate _{8}C_{2}, _{8}C_{1}, and _{8}C_{0} and then add _{8}C_{2}, _{8}C_{1}, and _{8}C_{0} together.

8!
2!(8 - 2)!

8!
2!(6)!

8×7×6!
2!(6)!

8×7
2

= 28
**Useful shortcuts to find combinations**

_{n}C_{1} = n and _{n}C_{0} = 1

Therefore, _{10}C_{1} = 10 and _{10}C_{0} = 1_{8}C_{2} + _{8}C_{1} + _{8}C_{0} = 28 + 10 + 1 = 39

The company has 39 ways to choose two or fewer candidates.

**Word problem #6**

A company has 20 male employees and 30 female employees. A grievance committee is to be established. If the committee will have **as many as** 3 male employees and **as many as** 2 female employees, how many ways can the committee be chosen?

**Solution**

The expression **as many as** makes the problem quite complex now since we now have all the following cases to consider.

Choose 3 males, 2 males, 1 male, or 0 male

Choose 2 females, 1 female, or 0 female.

Here is a complete list of all the different cases.

- 3 males and 2 females
- 3 males and 1 female
- 3 males and 0 female
- 2 males and 2 females
- 2 males and 1 female
- 2 males and 0 female
- 1 male and 2 females
- 1 male and 1 female
- 1 male and 0 female
- 0 male and 2 females
- 0 male and 1 female
- 0 male and 0 female

We only need to find _{20}C_{2}

20!
2!(18)!

20×19×18!
2(18)!

20×19
2

= 190
3 males and 2 females: _{20}C_{3} × _{30}C_{2} = 1140 × 435 = 495900 (done in **problem #4**)

3 males and 1 female: _{20}C_{3} × _{30}C_{1} = 1140 × 30 = 34200

3 males and 0 female: _{20}C_{3} × _{30}C_{0} = 1140 × 1 = 1140

2 males and 2 females: _{20}C_{2} × _{30}C_{2} = 190 × 435 = 82650

2 males and 1 female: _{20}C_{2} × _{30}C_{1} = 190 × 30 = 5700

2 males and 0 female: _{20}C_{2} × _{30}C_{0} = 190 × 1 = 190

1 male and 2 females: _{20}C_{1} × _{30}C_{2} = 20 × 435 = 8700

1 male and 1 female: _{20}C_{1} × _{30}C_{1} = 20 × 30 = 600

1 male and 0 female: _{20}C_{1} × _{30}C_{0} = 20 × 1 = 20

0 male and 2 females: _{20}C_{0} × _{30}C_{2} = 1 × 435 = 435

0 male and 1 female: _{20}C_{0} × _{30}C_{1} = 1 × 30 = 30

0 male and 0 female: _{20}C_{0} × _{30}C_{0} = 1 × 1 = 1

**Add everything:**

495900 + 34200 + 1140 + 82650 + 5700 + 190 + 8700 + 600 + 20 + 435 + 30 + 1 = 629566.

The number of ways to choose the committee is 629566