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Here's the question:

In an organization there are $80$ people, $40$ men and $40$ women. In how many ways can we choose, from those $80$ people, a $31$ member management so that there is a majority of women?

My answer was to first choose $16$ women (to assure the majority) out of the $40$ meaning $\binom{40}{16}$, and then choose the rest $15$ from the rest of all people meaning $\binom{64}{15}$. So the final answer is ${40 \choose 16} \cdot{64 \choose 15}$ but that isn't the right answer so I really don't know.. help? Thank you.


Note $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, the binomial coefficient.

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  • $\begingroup$ strange. $\binom{40}{16} \cdot \binom{64}{15}$ sounds right to me $\endgroup$ – gt6989b Jan 25 '16 at 23:02
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Jan 25 '16 at 23:23
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    $\begingroup$ @gt6989b The answer Noam obtained counts the same woman more than once when there are more than $16$ women. $\endgroup$ – N. F. Taussig Jan 25 '16 at 23:45
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How many 31 member committees can you make while ignoring the gender mix?

That would be $\binom{80}{31}$

What fraction of those will have a female majority? Think symmetry, since you have a lot of it in this case.

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The reason your solution gave the wrong answer is that you are sometimes counting the same committee more than once, e.g. if you:

  • choose women #1..#16 and then choose woman #17 + men #1..14

that's the same as

  • choose women #1..#15,woman#17 and then choose woman #16 + men #1..14

One productive way of thinking about the question is to ask "how many ways are there of choosing a committee with an equal number of men and women?", then think about the other committees: What proportion of them have more men than women? What proportion have more women than men?

Edit: that's not such a productive line of thought. See DJohnM's answer for where to go next.

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The correct calculation using your logic should go as follows: Remember we are looking for the number of committees where there is a majority of women.

Well, it could be the case that there 16 women and 15 men, and this should be $$\binom{40}{16}\binom{40}{15}.$$ But this is not the only case; it could be the case that there are 17 women and 14 men: $$\binom{40}{17}\binom{40}{14}.$$ Notice that since the composition/ratio of men and women are different, we are not counting the same committees.

Following this logic, we add up all the cases which is $$\sum_{k = 16}^{31}\binom{40}{k}\binom{40}{31-k}.$$

But this is not the ideal method/answer. You want to approach it as the other answer/post suggests.

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  • $\begingroup$ You mean $\sum_{k = 16}^{31}\binom{40}{k}\binom{40}{31-k}.$ $\endgroup$ – true blue anil Jan 26 '16 at 4:07
  • $\begingroup$ @trueblueanil Yes! Thank you!! $\endgroup$ – Em. Jan 26 '16 at 4:11

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