Exercise 6.2 — Question 22

Problem

In how many ways can a committee of 5 members be selected from 6 men and 4 women if the committee must contain at least 2 women?

Key Formula

The number of ways to choose $r$ items from $n$ distinct items (order does not matter) is:

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

Solution

We need a committee of 5 from 6 men and 4 women with at least 2 women.

"At least 2 women" means: exactly 2 women, exactly 3 women, or exactly 4 women.

Case 1: Exactly 2 women and 3 men

$\left(\genfrac{}{}{0ex}{}{4}{2}\right)\times \left(\genfrac{}{}{0ex}{}{6}{3}\right)=6\times 20=120$

Case 2: Exactly 3 women and 2 men

$\left(\genfrac{}{}{0ex}{}{4}{3}\right)\times \left(\genfrac{}{}{0ex}{}{6}{2}\right)=4\times 15=60$

Case 3: Exactly 4 women and 1 man

$\left(\genfrac{}{}{0ex}{}{4}{4}\right)\times \left(\genfrac{}{}{0ex}{}{6}{1}\right)=1\times 6=6$

Total

$120+60+6=\boxed{186}$

Key Concepts Used