Exercise 6.2 — Question 13

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?

Solution

We need to choose 3 men from 8 men AND 2 women from 6 women.

Using the combination formula: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

Step 1: Choose 3 men from 8: $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=\frac{8!}{3!\cdot 5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=56$

Step 2: Choose 2 women from 6: $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{2!\cdot 4!}=\frac{6\times 5}{2\times 1}=15$

Step 3: Apply the Fundamental Counting Principle (both selections must happen together): $\text{Total committees}=\left(\genfrac{}{}{0ex}{}{8}{3}\right)\times \left(\genfrac{}{}{0ex}{}{6}{2}\right)=56\times 15=840$

$\boxed{840\text{ committees}}$

Key Concept

When a selection problem has two independent groups and requires a fixed number from each, multiply the individual combinations: $\left(\genfrac{}{}{0ex}{}{m}{r_1}\right)\times \left(\genfrac{}{}{0ex}{}{n}{r_2}\right)$

This is because each choice from the first group can be paired with every choice from the second group.