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Given $m>n$ , In how many ways $ m$ men and $n$ women can seat in row for a photograph so that no two women are adjacent?

My effort : There are $m-1$ gaps if $m$ men are seated. Now we have to seat the women in the gaps and this can be done in $\binom{m-1}n$ ways. Now the answer should be $m!*\binom{m-1}n$

Can anyone verify please.


marked as duplicate by Marc van Leeuwen, Claude Leibovici, Ali Caglayan, Omnomnomnom, Venus Jan 21 '15 at 16:33

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    $\begingroup$ And for the verification; you counted the permutations among the men, but forgot to count the permutations among the women. Also women may be on the outside, so there are in fact $m+1$ "gaps" avaiable, not $m-1$. So replalce $m-1$ by $m+1$, and multiply your result by $n!$, giving $m!n!\binom{m+1}n$. $\endgroup$ – Marc van Leeuwen Jan 21 '15 at 15:40

For $m$ men we are having $m+1$ places available for $n$ women.Hence ways of seating arrangement of women =$P_n^{m+1}$ Now for $m$ we have $m!$ ways of seating them .Hence number of ways =$P_n^{m+1}m!$

  • $\begingroup$ What about the permutations of women? For each gender assignment ($m+1 \choose n$ of them) to the chairs, we have $n! m!$ ways of filling the chairs. $\endgroup$ – Mark Jan 21 '15 at 15:48
  • $\begingroup$ think about it i have justified it properly $\endgroup$ – Learnmore Jan 21 '15 at 15:49
  • $\begingroup$ @Mark Note that $P^n_k = k! \binom nk$. $\endgroup$ – AlexR Jan 21 '15 at 15:53
  • $\begingroup$ Oh thanks. For some reason I just assumed $P^n_k = {n\choose k}$ I'm not familiar with that notation. $\endgroup$ – Mark Jan 21 '15 at 15:58

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