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?
If you seat John first, he sits on the end with probability $\frac 27$ then Mary has $\frac 56$ chance not to sit next to him, or he sits in the middle with probability $\frac 57$ and Mary has $\frac 46$ chance not to sit next to him.
$$\frac 27 \cdot \frac 56+\frac 57 \cdot \frac46=\frac {30}{42}=\frac 57$$  

In the rest of your computation you are not considering order, so you shouldn't for JM either.
Easier to think about it by first calculating the probability that they WILL sit together

0 0 0 0 0 0 0  Seven seats

J 0 0 0 0 0 0  (1/6)

0 0 0 0 0 0 J  (1/6)  

0 0 0 J 0 0 0  (2/6) * 5

((2/6) * 5 + (1/6) * 2) / 7

Final answer:
1 - ((2/6) * 5 + (1/6) * 2) / 7) =
1 - ((10/6 + 2/6) / 7) =
1 - ((12/6) / 7) =
1 - (2/7) = 5/7