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The question is in how many ways we can order $m$ men and $n$ women in a row. I suggest it's a very simple one, but I'm confused by it :(

The answer is ${ m+n \choose n}$ = ${m+n \choose m}$

But I can't quite understand it. I'm not sure I get the task right, but isn't it in how many ways we can order them, no matter if man is next to woman or all men are together and so on, so it's the same as in how many ways we can order $(m+n)$ people which is $(m+n)!$

Or if all men should stand together and all women should stand together, isn't it $(m! + n!) * 2$, because we can order $m$ men in $m!$ ways and $n$ women in $n!$ ways and multiply it by two, because we can have the men first and then women or the opposite.

Please help me understand this, because it's one of the basic tasks and I wouldn't be able to understand all similar tasks after this :(

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    $\begingroup$ I agree with you. The natural interpretation is that women are distinguishable, as are men. If the question said $m$ identical red balls and $n$ identical black balls, then $\binom{m+n}{m}$ would be the correct answer. $\endgroup$ – André Nicolas Apr 21 '14 at 15:01
  • $\begingroup$ That is the natural way to think about people, but on seeing the question you have to ask what is significant about the fact the question distinguishes between men and women but otherwise does not distinguish between people. Then you can "naturally" conclude that it's because people are otherwise indistinguishable. $\endgroup$ – Steve Jessop Apr 21 '14 at 15:29
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    $\begingroup$ "Or if all men should stand together and all women should stand together, isn't it $(m!+n!)∗2$" -- actually that would be $m!*n!*2$, since each ordering of men could be combined with each ordering of women in 2 different ways. Assuming $m > 0$ and $n > 0$, that is. If one of them is $0$ then the 2 "different ways" are the same. Getting edge cases like that wrong is a common error in combinatorics. $\endgroup$ – Steve Jessop Apr 21 '14 at 15:32
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You do not make the difference between men and women.

You have $n+m$ people, with $n$ men. $\binom {m+n}n$ is the number of ways to decide where the men will be in the line (you do not say anything about which man in which place, just where men and women will be).

For instance, with 2 men and 2 women you get these possibilities:

MMWW
MWMW
MWWM
WMMW
WMWM
WWMM
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i think you understand the problem not correct. Here you do not have to distinguish between the individuals. You have only distinguish them by male or female.

So if all man stand together and all woman stand together, then you have only two oders. For example m=2,n=4

1.m,m,m,m,n,n

2.n,n,m,m,m,m

greetings,

calculus

Edit: Sorry, i´m too late.

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