$3$ women and $4$ men are standing in a line. If no two women may be adjacent to each other, how many distinct line-ups are there?
I'm not sure how to do this. I know $4! = 24$ and $3! = 6$. Where can I take it from there?
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asked Apr 11 '15 at 21:32
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1$\begingroup$ See this $\endgroup$ – IAmNoOne Apr 11 '15 at 21:42
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The $4$ men stand in a line giving $4!=24$ possibilities. There are $5$ places where a woman could go. One to the left of the men, one to the right, and three in the middle between some of the men. No two of the women may be placed in the same of these $5$ places. So there are $5*4*3=60$ possible combinations permutations for the women among the men, giving a total of $24*60=1440$ possible lineups.
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$\begingroup$ Your answer is correct. However, the word combination ordinarily suggests an unordered selection. Here, you are making an ordered selection, that is, a permutation. $\endgroup$ – N. F. Taussig Apr 12 '15 at 21:55
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$\begingroup$ Thanks for the comment. I'll remember the distinction now. $\endgroup$ – Josh B. Apr 13 '15 at 2:40
