With probability
In how many ways can 5 married couples sit in a row if no 2 women sit next to each-other?
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This is just a combinatorics problem, not probability. I'm assuming you mean heterosexual couples.
No 2 women sitting next to each other means that there has to be at least one man between two women. So, there are six available spots for women:
_ M _ M _ M _ M _ M _
With five available women, there are (6 choose 5) ways to pick the spots. Assuming that the people are distinct, there are 5! ways to order the men and 5! ways to order the women. So, the answer is
5! $\cdot$ 5! $\cdot$ (6 choose 5)
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when I did that I got 86400? – Victor Taylor Perry May 09 '13 at 18:05
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It seems correct to me. – Narut Sereewattanawoot May 09 '13 at 18:07
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You were correct! – Victor Taylor Perry May 09 '13 at 18:09
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can you answer it now if it was 4 married couples? – Victor Taylor Perry May 09 '13 at 18:22
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4! $\cdot$ 4! $\cdot$ (5 choose 4) – Narut Sereewattanawoot May 09 '13 at 18:23
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I think is like
_ M _ M _ M _ M _ M _
so women has 6 seats and they are 5. and 5 men has 5 seats. so answer will be $6P5$*$5!$
iostream007
- 4,529
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Interesting... I've got 3 different ways to approach the problem...all seem right... – Victor Taylor Perry May 09 '13 at 18:06
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2@VictorTaylorPerry: This one considers different orders of one sex to be the same while different orders of the other are different. It doesn't seem consistent. – Ross Millikan May 09 '13 at 18:20
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You need to multiply by
5!for all possible ordering of men and also by another5!for all possible ordering of women. – Petr May 10 '13 at 14:02