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We have $n$ men and $m$ women and $n\geq m $. In how many ways can they stand in circle that no two women stand next to each other in terms of $n$ and $m$?

Thanks.

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  • $\begingroup$ I think that the total number of ways to stand n men and m women in cycle is $(n+m-1)!$. Suppose two women stand next to each other and then number of ways $2(^mC_2)(n+m-2)!$. So the answer to your question is $(n+m-1)!-2(^mC_2)(n+m-2)!$. $\endgroup$ – hemu Nov 19 '17 at 9:49
  • $\begingroup$ #shrinit I think, there are some ways substracted more times. If 3 women stand next to each other, one bad permutation will be substracted 2 times. $\endgroup$ – Přemysl Šťastný Nov 19 '17 at 10:22
  • $\begingroup$ Yes, thank you. $\endgroup$ – hemu Nov 19 '17 at 10:29
  • $\begingroup$ Yes, thank you. You already got the answer. But if you add (in my previous answer) (number of ways when 3 women stand next to each other - number of ways when 4 women stand next to each other and continue it until m women stand next to each other. I think then it will be the answer but not useful one. Once again thank you for pointing that out $\endgroup$ – hemu Nov 19 '17 at 10:38
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First deal with our men. They can be arranged in the circle in $(n-1)!$ ways. Now for each of these ways, we have $n$ positions left for our women. As there are $m$ women, we can choose a position for them in $\binom{n}{m}$ ways, and arranged in $m!$ ways. In all:

$$\text{No of ways = }(n-1)! \binom{n}{m} (m!)$$

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If all women stand together, then there are $(m-1)!$ ways. Between two of them, there is a man (minimal). So, there are $(m-1)! P^{n}_{m}$ ways. Correct me if I wrong.

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  • $\begingroup$ What does $P_m^n$ mean, please? $\endgroup$ – Přemysl Šťastný Nov 19 '17 at 10:04
  • $\begingroup$ If all women stand together, there are m blank places(it's cyclic, isn't?). Meanwhile, there are n men in m blank places. We get permutation $P^n_m.$ $\endgroup$ – Maurten Erik Nov 19 '17 at 10:19
  • $\begingroup$ I think you count permutations in which there are no men between 2 women too. $\endgroup$ – Přemysl Šťastný Nov 19 '17 at 11:02

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