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In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?

Attempt:

As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$

The answer is $371$. Where is my error?

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  • $\begingroup$ I don't think any of the answers answer my question. I wonder why I don't get the correct answer following my train of thought. If I guaranteed the combination of at least $2$ women in the choices, I would multiply it by the combination of the rest of the available people: $4$ between $9$, counting the $7$ men and $2$ women. $\endgroup$ Nov 27, 2021 at 23:17
  • $\begingroup$ @Behemooth Your error is that you are considering only the case where $2$ women are picked when the problem asks for "at least", and there are only $7$ men to choose from, not $9$. Fixing that, you get my answer. $\endgroup$
    – Evariste
    Nov 27, 2021 at 23:21
  • $\begingroup$ @Evariste There are $7$ men to choose from + $2$ women remaining unchosen. There will be a combination of men + women in the last 4 choices. $\endgroup$ Nov 27, 2021 at 23:23

4 Answers 4

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If you choose six people and at least two women are selected, then either two women and four men, three women and three men, or four women and two men are selected. The number of ways of selecting $k$ women and $6 - k$ men from four women and seven men is $$\binom{4}{k}\binom{7}{6 - k}$$ Hence, the number of selections with at least two women is $$\sum_{k = 2}^{4} \binom{4}{k}\binom{7}{6 - k} = \binom{4}{2}\binom{7}{4} + \binom{4}{3}\binom{7}{3} + \binom{4}{4}\binom{7}{2} = 371$$

You counted each selection with more than two women multiple times.

Your method counts each selection with three women three times, once for each of the $\binom{3}{2}$ ways of designating two of the three selected women as the two women in the group, and each selection with all four women six times, once for each of the $\binom{4}{2}$ ways of designating two of the four women as the four women in the group.

For instance, suppose the women are Anne, Brenda, Charlotte, and Diana and the men are Edward, Frank, George, Henry, Ivan, Jeffrey, and Karl. You count the selection of Anne, Brenda, Charlotte, Edward, Frank, and George three times.

$$\begin{array}{l | l} \text{designated women} & \text{additional people}\\ \hline \text{Anne, Brenda} & \text{Charlotte, Edward, Frank, George}\\ \text{Anne, Charlotte} & \text{Brenda, Edward, Frank, George}\\ \text{Brenda, Charlotte} & \text{Anne, Edward, Frank, George} \end{array} $$

Similarly, your method counts the selection of Anne, Brenda, Charlotte, Diana, Edward, and Frank six times.

$$\begin{array}{l | l} \text{designated women} & \text{additional people}\\ \hline \text{Anne, Brenda} & \text{Charlotte, Diana, Edward, Frank}\\ \text{Anne, Charlotte} & \text{Brenda, Diana, Edward, Frank}\\ \text{Anne, Diana} & \text{Brenda, Charlotte, Edward, Frank}\\ \text{Brenda, Charlotte} & \text{Anne, Diana, Edward, Frank}\\ \text{Brenda, Diana} & \text{Anne, Charlotte, Edward, Frank}\\ \text{Charlotte, Diana} & \text{Anne, Brenda, Edward, Frank} \end{array} $$

Notice that $$\binom{4}{2}\binom{7}{4} + \color{red}{\binom{3}{2}}\binom{4}{3}\binom{7}{3} + \color{red}{\binom{4}{2}}\binom{4}{4}\binom{7}{2} = \color{red}{756}$$

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  • $\begingroup$ +0.95 for the excellent explanation and catching the OP's intentions right after their comment on the rest of the answers, and +0.05 for the choice of names having first letters from $A$ to $K$, always fun to see such examples. $\endgroup$ Nov 28, 2021 at 8:45
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First choosing $W_1$ and $W_2$ to satisfy the requirement, then choosing $W_3$ among the other four people gives the same group as first choosing $W_1$ and $W_3$, then choosing $W_2$ among the other four. Because of this, your method is counting many groups multiple times.

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  • $\begingroup$ As far as I know of the definition of "combination", it doesn't care about the order of choices. Why is the combination counting more than once? $\endgroup$ Nov 27, 2021 at 23:25
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The number of ways is

$$\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2} =6\times35+4\times35+21 = 371$$

"At least two women" means either $2$, $3$ or $4$.

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Consider three cases:

  • choosing 2 women and 4 men
  • choosing 3 women and 3 men
  • choosing 4 women and 2 men

So the total number of arrangements is: $\binom{4}{2} \times \binom{7}{4}+ \binom{4}{3} \times \binom{7}{3}+ \binom{4}{4} \times \binom{7}{2}=371$

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