Question

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

Answer 1

is that 21

Answer 2

7C2

Answer 3

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

Answer 4

got it?

Answer 5

5040-2*720/5040

Answer 6

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

Answer 7

thnx guys

Answer 8

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.

Answer 9

thnx queelius