The question is find how many ways we may select a committee of $5$ members from $6$ men and $4$ women such that at least two women are included.
I know that the standard approach to this is to take cases like
- $2$ women, $3$ men
- $3$ women, $2$ men
- $4$ women, $1$ man
And then we can proceed. Doing so the answer arrives to be $186$ ways.
My doubt is why can't we write ${}^4 C_2 \times {}^8 C_3$.
$2$ women are selected first then out of $6$ men and remaining $2$ women (total $8$) any $3$ are selected. Why is this wrong?
Suppose we label the women $W_1,W_2,W_3,W_4$ and the men $M_1,M_2,M_3,M_4,M_5,M_6$.
Suppose, under your scheme, we choose two women first, and happen to pick $\color{blue}{W_1},\color{blue}{W_2}$, and then pick the remaining three people and happen to pick $\color{red}{W_3},\color{red}{M_1},\color{red}{M_2}$, so that
$$\text{our committee is } \{\color{blue}{W_1},\color{blue}{W_2},\color{red}{W_3},\color{red}{M_1},\color{red}{M_2}\}$$
But suppose we choose another committee. It is possible we pick $\color{blue}{W_2},\color{blue}{W_3}$ first, and then the remainder of the committee is $\color{red}{W_1},\color{red}{M_1},\color{red}{M_2}$. Then
$$\text{our committee is } \{\color{red}{W_1},\color{blue}{W_2},\color{blue}{W_3},\color{red}{M_1},\color{red}{M_2}\}$$
again! We have picked the same committee in two different ways. (I chose to use colors to distinguish the first choice of women in blue, and the remaining three people in red, but notice the committee still has the exact same composition, by the exact same people.)
That is to say, your method results in overcounting (supported by the fact that your answer is $336>186$).