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How many different ways are there for eight men and five women to stand in a line so that no two women stand next to each other?

I have the answer as $P(8,8) \cdot P(9,5)$

But how is $P(9,5)$ possible?

Please, anyone explain this to me.

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closed as off-topic by Namaste, Isaac Browne, Leucippus, Saad, Shailesh Apr 24 '18 at 2:15

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We can line up the eight men in $P(8, 8)$ ways. This creates nine spaces, seven between successive males and two at the ends of the row. $$\square M \square M \square M \square M \square M \square M \square M \square M \square$$ To ensure that the women are separated, we must place the five women in five of those nine spaces, which can be done in $P(9, 5)$ ways.

Another way to think about it: To ensure that the women are separated, we choose five of the nine spaces in which to place a woman, which can be done in $\binom{9}{5}$ ways. The five women can then be arranged in those spaces in $5!$ ways. Hence, the number of ways the women can be placed in the line once the men have been arranged is $$\binom{9}{5}5! = \frac{9!}{5!4!} \cdot 5! = \frac{9!}{4!} = \frac{9!}{(9 - 5)!} = P(9, 5)$$

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You have $9$ boxes (position in the male line). How many ways do you have to place 5 balls (women) in distinct boxes (such that there is at most 1 ball per box).

i.e how many way do you have to pick $5$ distinct numbers (the box numbers for the 5 females) from $9$ numbers (all the possible position) ? $9\choose5$

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  • $\begingroup$ The OP might benefit from a diagram that shows where you are placing the balls relative to the males. $\endgroup$ – N. F. Taussig Apr 23 '18 at 17:44

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