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To a certain conference, each firm can send two employee representatives, on the condition that one of them is a male and the other a female. If 15 firms were represented in this conference, what is the probability that no two females are seated next to each other? (Assume that all 30 members are seated in a single row of seats) A] 1/2 B] 14!/30! C] 15!/30! D] 28/30!

For above question I am trying following: 15 men can be arranged in 15! ways. There are 16 spots among men (-M-M-.....-M-). 16 spots can be taken by 15 women in 16! ways. My answer is (15!*16!)/30!. But it's not listed in answer choices. Am I missing something or answers are incorrect?

My approach is similar to approach in question in below link. How many ways are there for 10 women and six men to stand in a line

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    $\begingroup$ Your analysis is correct. $\endgroup$ – André Nicolas Jul 23 '15 at 15:43
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    $\begingroup$ You are welcome. I thought for a while about what strange interpretation of the wording could make one of the proposed answers correct, but could not come up with one. I would probably instead say there are $\binom{30}{15}$ equally likely ways to choose the positions for the women, of which $\binom{16}{15}$ are "favourable." Divide, same answer. $\endgroup$ – André Nicolas Jul 23 '15 at 16:18
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$\bf{My\; Solution::}$ If we can represent man by $\bf{M}$ and Women by $\bf{W}$, Then we use Gap Method.

So Arrangements as $$_M_M_M_M_M_M_M_M_M_M_M_M_M_M_M_$$

Above $15$ man can be arrange as $15!$ (Here Man and women all are Different)

Now We can Arrange $15$ Women in These $16$ Gap, Which can be done by $\displaystyle \binom{16}{15}\times 15! = 16!$

So Total no. of Arrangement in which no $2$ woman sit together is $\displaystyle 15! \times 16!$

And Total Probability $\displaystyle = \frac{\bf{Favourable \; cases}}{\bf{Total \; Cases}} = \frac{15!\times 16!}{30!}$

So I Think Your answer is Right.

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I think the answer is wrong because it doesn't take in account that two men can seat side by side (if you have tow women on te exterior) : WMWMWM..WMMW

Something like ${30*15!*15!\over30!}$

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  • $\begingroup$ The proposed solution is correct. Note that if no two women sit next to each other, they must fill one of the $14$ spaces between adjacent men or the two spaces at the ends. If both spaces at the ends are filled, one of the spaces between adjacent men does not contain a woman, so those two men do sit next to each other. $\endgroup$ – N. F. Taussig Jul 23 '15 at 19:45

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