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?

Have John and Mary "reserve" a pair of seats.  There are ${7\choose2}=21$ pairs possible, $6$ of which are side by side.  So if they make a reservation at random, the probability they'll wind up sitting apart is
$$1-{6\over{7\choose2}}=1-{6\over21}={5\over7}$$
Alternatively, have John and the five others go stand in a row near the chairs. Then, before anyone sits down, have Mary come join them, inserting herself either between two people or at one of the two ends.  There are $7$ places Mary can insert herself, only $2$ of which are next to John, so the probability Mary and John wind up sitting apart is $5/7$.  (This is essentially the same answer at true blue anil's, mostly just expressed in story form.)
