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hager
The 7 members of a chess club line up for a picture. Determine the probability that Jordan and Ryan are not beside each other.
$$\frac{7!-2!6!}{7!}=1-\frac{2!6!}{7!}=1-\frac27=\frac57$$
Others have already mentioned the solution but I thought I would explain how you get it. The way I solved it was by first finding out the probability of them being next to each other. So, we have 7 places. If they're next to each other, we have the following cases: Let A be Jordan and B by Ryan. A B _ _ _ _ _ B A _ _ _ _ _ _ _ A B _ _ _ _ _ B A _ _ _ And so on. In total, we have 12 configurations of this. In each of the blank positions, we do a simple permutation of the remaining people, i.e., 5!. So, we have 12 * 5! ways of arranging them, which equals 2 * 6! ways. There are 7! ways in total, without the constraint. So, the probability they are next to each other is: 2 * 6! / 7! = 2 / 7 The probability that they are not next to each other is 1 - 2 / 7, then, which is 5 / 7.