# In how many ways can you form a committee of three from a set of $10$ men and $8$ women, such that there is at least one woman in the committee?

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?

• You count WWM, WMW, WWW cases multiple times maybe? Jun 15 '20 at 7:34

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.
$$C(18,3)-C(10,3)=696$$.
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: $$\ \text{Possible Ways} = \text{Total Ways} - \text{Ways with only men}\\ \text{Possible Ways} = \left(\begin{smallmatrix} 18 \\ 3 \end{smallmatrix}\right) - \left(\begin{smallmatrix} 10 \\ 3 \end{smallmatrix}\right)=696$$