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five friends including Bilyana and Bojana are sitting in a row in a theatre determine the probability that they are not sitting together. This is part of a homework assignment. I don't even know where to start with this.

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  • $\begingroup$ The trick to the problem is determining how many ways the two friends can sit together. $\endgroup$ – Chris K Dec 1 '13 at 0:49
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We will put on a couple of the seats a sign saying "Only to be used by Bilyana or by Bojana. Anyone else, Bulgarian or not, keep off."

There are $\binom{5}{2}$ equally likely ways to select the two seats. Exactly $4$ of these choices leave us with the two B's sitting together. Thus our probability is $1-\frac{4}{\binom{5}{2}}$.

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Hint:

Think of sitting Bilyana first, Bojana second and the rest next. How many ways of choosing a seat is there for Bilyana? and then for Bojana? (consider there's a different answer to this if Bilyana is (is not) seating at one end) What about the others next?

Work on this and if you still can't find your answer, update the question including your work so far.

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