# A committee of $4$ has to be formed from $6$ men and $4$ women must contain at least one woman

combinationscombinatorics

From a group of $$6$$ men and $$4$$ women a committee of $$4$$ persons is to be formed in such a way that committee has at least one woman.

I tried to divide this work into 2 tasks-

1. Selecting $$1$$ woman out of $$4$$ to fulfill the requirement
2. Then selecting $$3$$ members from rest of the people

So,the answer which I get from this is,

$$9C3 × 4C1 = \frac{9 \cdot 8 \cdot 7}{6} \cdot 4 = 336$$

But this is not the correct answer. What am I doing wrong here??

Your method counts each committee with exactly $$k$$ women $$k$$ times, once for each way you could designate one of those $$k$$ women as the woman on the committee.
There are $$\binom{4}{k}\binom{6}{4 - k}$$ ways to select a committee of four people with exactly $$k$$ women and $$4 - k$$ men, so the number of four-person committees with exactly $$k$$ women is $$\sum_{k = 1}^{4} \binom{4}{k}\binom{6}{4 - k} = \binom{4}{1}\binom{6}{3} + \binom{4}{2}\binom{6}{2} + \binom{4}{3}\binom{6}{1} + \binom{4}{4}\binom{6}{0} = 195$$ which, as indicated in the comments of Math Lover and Thomas Andrews, can be found more simply by subtracting the number of committees with only men from the number of four-person committees. $$\binom{10}{4} - \binom{6}{4} = 195$$
$$\begin{array}{c | c} \text{designated woman} & \text{additional people}\\ \hline \text{Angela} & \text{Barbara, Charles, Douglas}\\ \text{Barbara} & \text{Angela, Charles, Douglas} \end{array}$$
Similarly, your method counts each committee with three women three times and the committee with all four women four times. Notice that $$\binom{4}{1}\binom{6}{3} + \color{red}{\binom{2}{1}}\binom{4}{2}\binom{6}{2} + \color{red}{\binom{3}{1}}\binom{4}{3}\binom{6}{1} + \color{red}{\binom{4}{1}}\binom{4}{4}\binom{6}{0} = 336$$ where the factors in red denote the number of ways you could designate one of the $$k$$ women on the committee as the woman on the committee.