Why is the following method incorrect when deciding on the possible choices?

You are counting each selection with two women twice and each selection with three women three times, once for each way you could have designated one of the women who is selected as the woman you have selected and the remaining woman or women as additional people.

Suppose you select two women, Abigail and Beatrice, and one man, Charles. You count this selection twice:

\begin{array}{l l} \text{woman} & \text{additional people}\\ \hline \text{Abigail} & \text{Beatrice, Charles}\\ \text{Beatrice} & \text{Abigail, Charles} \end{array}

Suppose you select three women, Abigail, Beatrice, and Charlotte. You count this selection three times:

\begin{array}{l l} \text{woman} & \text{additional people}\\ \hline \text{Abigail} & \text{Beatrice, Charlotte}\\ \text{Beatrice} & \text{Abigail, Charlotte}\\ \text{Charlotte} & \text{Abigail, Beatrice} \end{array}

Notice that $$\binom{3}{1}\binom{6}{2} + \color{red}{\binom{2}{1}}\binom{3}{2}\binom{6}{1} + \color{red}{\binom{3}{1}}\binom{3}{3} = 84$$ where the factors in red are the number of ways you could have designated one of the women you have selected as the woman you have selected.


There is a much simpler approach. "At least one woman" is the opposite of "zero women" or "all men" from the total, so subtracting from the total gives us:

$${9 \choose 3} - {6 \choose 3} = 64$$

If you are interested in another subtractive approach, following from N. F. Taussig's answer, you can again subtract from ${3 \choose 1} \cdot {8 \choose 2} = 84$.

The cases where there are two women and one man (Abigail, Beatrice, Charles) are counted $2$ times, when they should only be counted $1$ time. Therefore, you need to subtract the number of combinations of this type $1$ time. This is just ${3 \choose 2} \cdot 6 = 18$.

Similarly, the cases where there are three women are counted $3$ times, when they should only be counted $1$ time, as there is only one combination possible with all $3$ women. Therefore, the number of combinations should be subtracted by $2$.

Hence $84 - 18 - 2 = 64$, which is the correct answer.

Tags:

Combinatorics

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