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

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??

• @MathLover the answer given doesn’t count only one-woman committees. Rather the error is counting some committees more than once. Oct 3 '21 at 4:04
• OK I see now what you are doing. Sorry for previous comment. Your work has duplicates. Please find ways to form a committee with no women on it and then subtract from ${10 \choose 4}$ Oct 3 '21 at 4:05
• The error is that you are counting some committees more than once. If the committee has two women, you get that by selecting either woman first, then selecting the other three. Oct 3 '21 at 4:06
• The trick here is to count all possible committees with any number of women, then subtract the committees of only men. Oct 3 '21 at 4:07
• @ThomasAndrews thanks. yes I noticed later. Oct 3 '21 at 4:09

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.