A committee of 5 members is to be formed from 6 men and 4 women such that it contains at least one woman. How many such committees can be formed?

Solution

1. The total number of ways to choose 5 members from 10 people (6 men and 4 women) is given by $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{10}{5}\right)& =\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}\\ & =252\end{array}$$.
2. The number of ways to form a committee with no women is equivalent to choosing all 5 members from the 6 men, which is $\left(\genfrac{}{}{0ex}{}{6}{5}\right)=6$.
3. Subtract the committees with no women from the total number of committees to find those with at least one woman: $252 - 6 = 246$.
4. Hence the answer is (A).