The 7 members of a chess club line up for a picture. Determine the probability that Jordan and Ryan are not beside each other.

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- anonymous

- schrodinger

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- anonymous

is that 21

- anonymous

7C2

- anonymous

(7! - 2!*6!) /7!

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- anonymous

got it?

- anonymous

5040-2*720/5040

- anonymous

$$\frac{7!-2!6!}{7!}=1-\frac{2!6!}{7!}=1-\frac27=\frac57$$

- anonymous

thnx guys

- queelius

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.

- anonymous

thnx queelius

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