7 friends are going to the cinema. They will be sitting in a row with 7 seats. What is the probability that John and Mary don't sit together? 
To watch a movie, John, Mary and 5 friends will sit randomly in a row with 7 seats. What is the probability John and Mary won't sit together?
$$(\mathbf A)\ \frac{2\times5!}{7!}\qquad(\mathbf B)\ \frac{5!}{7!}\qquad(\mathbf C)\ \frac27\qquad(\mathbf D)\ \frac57$$

I did:
$$1-\left(6\cdot 2\cdot\left(\frac{2}{7}\cdot\frac{1}{6}\right)\right) = \frac{3}{7}$$
But my book states the solution is D). I tried not multiplying by 2 and I get D), however I don't know exactly why the 2 is wrong.
You can make 2 permutations with Mary(M) and John(J), MJ and JM.
Then if you imagine the 2 of them as a block of 2 seats they can sit in $^6C_1=6$ places.
Why doesn't my book count those 2 permutations of JM and MJ?

See the total ways are $7! $ now let $jm $ be one guy (not biologically) just assume. So now we have total $1+5=6$  ways. We can now arrange these as $6! $ and these two persons can be arranged within themselves in $2! $ thus total ways where they sit together are $2!.6! $hence probability that they wont sit together$=\frac {7!-2!6!}{7!}=1-\frac {2}{7}=\frac {5}{7} $
