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If we have a group of $10$ men, and $4$ women, and we want to separate these $14$ people to $2$ groups of $7$ such that each group has at least $1$ women, in how many different ways can we achieve this?

My solution is: We can look at just one group, since that once we chose for the first group, we automatically have chosen for the second group, so we have $14$ people and we want to choose $7$ people such that there is at least $1$ women, so we either have $1$ women and $6$ men, or $2$ women and $5$ men, or $3$ women and $4$ men. That's:

$\binom{4}{1}\binom{10}{6}+\binom{4}{2}\binom{10}{5}+\binom{4}{3}\binom{10}{4}=3192$

So I'd say we have $3192$ ways to do this. However, the booklet says the correct answer is half of that,$1596$. I seem to be counting double. Where is my mistake?

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    $\begingroup$ You have counted the number of ways to divide the group into two teams of $7$, one to wear blue uniforms and the other to wear red. You are asked for the number of ways to divide the group into two teams not wearing clothes. $\endgroup$ – André Nicolas Jan 18 '15 at 23:00
  • $\begingroup$ @andre what about the middle term ? It is symmetric. Were those ways also counted double ? $\endgroup$ – Peter Jan 18 '15 at 23:01
  • $\begingroup$ Yes, the $\binom{4}{2}\binom{10}{5}$ counts the number of ways to choose $2$ women and $5$ men to wear blue. $\endgroup$ – André Nicolas Jan 18 '15 at 23:04
  • $\begingroup$ But this coincides with the same with red clothes. I still do not see why these ways were counted double. $\endgroup$ – Peter Jan 18 '15 at 23:08
  • $\begingroup$ Let's forget about the men, and look at the simpler problem of dividing a group of $4$ women, A, B, C, D into two teams of two each. We can list the possibilities easily: on A's team B, C, or D, now everything is determined, $3$ ways. But $\binom{4}{2}=6$. $\endgroup$ – André Nicolas Jan 18 '15 at 23:26
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Hint:
Suppose you have the people in a line, the women and the man. Then, you label each person with a number $1$ or $0$. $0$ for the first group and $1$ for the second one. So $\color{pink}{1000}\color{blue}{1111110000}$ is one valid labeling (where pink ones are women and blue ones are men). But suppose you have this labeling too $\color{pink}{0111}\color{blue}{0000001111}$.
What is the difference between those two forms of grouping them?

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  • $\begingroup$ There is no difference. But I took that into consideration, since I only counted for 1 team. if I have 10 men and 4 women, and I need to separate them to 2 groups (lets call them $A$ and $B$), the way to choose members for group $A$ such that we have at least 1 woman is $\binom{4}{1}\binom{10}{6}+\binom{4}{2}\binom{10}{5}+\binom{4}{3}\binom{10}{4}$.I didn't even mention group $B$ there. $\endgroup$ – Oria Gruber Jan 18 '15 at 23:07
  • $\begingroup$ No, you did not because you are giving them an order by saying that you just take 1 team into consideration. Those two labelings are the same for your problem because you do not care if the first group is labeled by a 1 or a 0. $\endgroup$ – Phicar Jan 18 '15 at 23:09
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    $\begingroup$ Oh that's right!!Ok, I get it now. $\endgroup$ – Oria Gruber Jan 18 '15 at 23:10
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Your flaw is, that you counted how one group is assembled. However, for each group-constellation you count, you in fact have two, since the rest is another possible group.

Simple example: divide the two letters $A$ and $B$. Looking at one letter only, you have two choices: either $A$ or $B$. But think again. How many ways to separate them are there? ;)

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