<s>[INST] 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-

*

*Selecting $1$ woman out of $4$ to fulfill the requirement

*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??
 [/INST] 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.
$$\binom{10}{4} - \binom{6}{4} = 195$$
Your method counts each committee with two women twice.  For instance, if the women Angela and Barbara and the men Charles and Douglas are chosen to serve on the committee, your method counts this committee twice:
$$
\begin{array}{c | c}
\text{designated woman} & \text{additional people}\\ \hline
\text{Angela} & \text{Barbara, Charles, Douglas}\\
\text{Barbara} & \text{Angela, Charles, Douglas}
\end{array}
$$
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.
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