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My textbook employs a brute force method: add the number of committees that could be formed with one woman, two women, and three women in them. Then, the total number of such committees will be: $$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 2 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 2 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 1 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 3 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 0 \end{smallmatrix}\right) = 360 + 280 + 56 = 696$$ (The number of ways of choosing one woman from eight times that of choosing two men from ten + ...)

I had solved the question with this reasoning: You can choose one woman from eight as a member of the committee, and for the remaining two positions, you could choose either a man or a woman. This is equivalent to choosing two people from $(10 + 8) - 1 = 17$. Thus, the number of possible ways to form such a committee is: $$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 17 \\ 2 \end{smallmatrix}\right) = 8 \cdot 136 = 1088$$

What mistake have I made?

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  • $\begingroup$ You count WWM, WMW, WWW cases multiple times maybe? $\endgroup$ Jun 15 '20 at 7:34
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Suppose $w_1$ was the first woman chosen among possible $8$, then say $m_1,w_3$ were chosen. So your committee is $w_1,m_1,w_3$.

In your way of counting: what if $w_3$ is chosen as the first woman, followed by $m_1, w_1$? Then too the committee is same as before ($w_3,m_1,w_1$) but you are counting this as a different committee.

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Anurag has put a finger on it. Your way of counting implicitly makes order of selection matter. Here's a check though. The number of committees with at least one woman should be the total number of all committees less the number that contain only men:

$$C(18,3)-C(10,3)=696$$.

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You are recounting! For simplicity consider there are only three people. $W_1, W_2, M_1$. Suppose you count the same way. You will get number way as $\left(\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right) \times \left(\begin{smallmatrix} 2 \\ 2 \end{smallmatrix}\right)=2$. But are there two ways of forming the committee? No! You just recounted $W_1 (W_2 M_1)$ and $W_2 (W_1 M_1)$. The easier way could be: $$\ \text{Possible Ways} = \text{Total Ways} - \text{Ways with only men}\\ \text{Possible Ways} = \left(\begin{smallmatrix} 18 \\ 3 \end{smallmatrix}\right) - \left(\begin{smallmatrix} 10 \\ 3 \end{smallmatrix}\right)=696 $$

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