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Hi I have a question:

We have 4 couples sitting in a row side by side. In how many ways they can sit so at least 1 couple doesn't sit together? I was thinking about doing (8 choose 4 ) - ( 8 choose 3) but it doesn't make sense. Thanks for any help!

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    Your approach is good, but you need to think a bit more about this question: how many ways can they be seated so that both couples sit together? – 2017-02-15
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    Hmm for all couples seating together: 4!*2!*2!*2!*2!, those would be all the arrangements. But not sure how to split up one couple now – 2017-02-15
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    OR (4 choose 3)2!2!2! + (8 choose 2 ) – 2017-02-15

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Total number of arrangements which have all couples sitting together is $4! \cdot 2! \cdot 2! \cdot 2! \cdot 2! = 384$.

Total number of arrangements is $8! = 40320$.

Total number of arrangements where not all couples sit together is $1 - 384/40320 = 0.99048$

I'd post this as a comment but 50 rep.

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    We want the number, not the probability, but otherwise good answer. Remember to use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) when formatting math. – 2017-02-15