Exercise 6.2 — Question 17

Problem

How many committees of 5 members can be formed from 8 men and 6 women if the committee must consist of 3 men and 2 women?

Key Concept: Combinations

A combination is a selection of objects where order does not matter.

The combination formula is:

$C(n,r)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

When a selection problem has two independent groups, multiply the combinations from each group.

Solution

Step 1: Choose 3 men from 8 men.

$C(8,3)=\frac{8!}{3!\cdot 5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=56$

Step 2: Choose 2 women from 6 women.

$C(6,2)=\frac{6!}{2!\cdot 4!}=\frac{6\times 5}{2\times 1}=15$

Step 3: Apply the Fundamental Counting Principle — multiply the two results.

$\text{Total committees}=C(8,3)\times C(6,2)=56\times 15=840$

Answer: $\boxed{840}$ committees can be formed.

Why Multiply?

For each of the 56 ways to choose the men, there are 15 independent ways to choose the women. By the multiplication principle, the total number of ways is $56\times 15=840$.