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?
 A: The number of ways with MJ or JM is $2 \cdot \,^6C_1 \cdot \,^5P_5$.
The total number of ways is $\,^7P_7$.
Hence the required probability is $$1 -\frac{2 \cdot 6 \cdot 5!}{7!} = 1 -\frac{2 \cdot 6!}{7!} = 1 - \frac{2}{7} = \boxed{\frac{5}{7}}$$
A: $\dfrac{6\cdot5}{6\cdot7} = \dfrac57\quad$ Logic ?
Depicting the $2$ "specials" and the $5$ "others" as red /white balls respectively, 
The first red can always be placed anywhere in $6$ ways with the whites,
 e.g. ${\Large\circ\circ\circ\circ\color{red}{\bullet}\circ}$
but wherever you place it, the second red has only $5$ authorised places
e.g. $\;{\Large\uparrow\circ\uparrow\circ\uparrow\circ\uparrow\circ\color{red}\bullet\circ\uparrow}\;$ against $7$ unconstrained places,
thus $Pr = \dfrac{6\cdot5}{6\cdot7} = \dfrac57$
A: 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.
A: 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} $
A: 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.)
A: I just spotted the mistake. I should have done either:
$$1-(6\cdot(\frac{2}{7}\cdot\frac{1}{6})) = \frac{5}{7}$$
or 
$$1-(6\cdot2\cdot(\frac{1}{7}\cdot\frac{1}{6})) = \frac{5}{7}$$
because $2\cdot(\frac{1}{7}\cdot\frac{1}{6}) = \frac{2}{7}\cdot\frac{1}{6}$
A: I thought about the following: We have to find the total number of arrangements without restrictions and then we find the total number of arrangements where John and Mary sit together. This is the probability they sit together. This gives: JM together: $=2×6!$, No res. $=7!$
$\frac{2×6!}{7!}$ gives $\frac{2}{7}$ thus the probability they do not sit together is $\frac{5}{7}$
A: So the overall number of sitting options is $7!$, I think nobody can argue over this.
WLOG let's say we're sitting everybody left to right: 
Let MJ (Mary sitting left to John) be a sitting block, then the number of possible sitting options on this case is $6!$
Let JM be the complementary block (i.e Mary sitting right to John), these are totally different $6!$ options.
Together we've got $2\cdot 6!$ options where Mary and John sit together during the movie, which gives us probability of $\frac{2\cdot6!}{7!}=\frac{2\cdot6!}{7\cdot6!}=\frac{2}{7}$ for the event of Mary and John making out.
The complementary probability (i.e 'wait patiently till you get to your room') will be $1-\frac{2}{7}=\frac{5}{7}$. Voilà!
A: The reason your solution didn't quite work is that you assumed that there is only one seat that the John can sit on if Mary sits first, or vice versa if John sits first. This is shown by the product $\frac{2}{7}\cdot \frac{1}{6}$ in your solution. This doesn't work, because there are cases where there are two seats next to John in which Mary can sit on. You must use casework if you approach the problem the way you did. 
The best way to do this is to consider the total number of permutations that are successful for both the desired cases and total cases rather than to approach it by straight probabilities. 
Doing this, there are $7!$ ways for everyone to sit however they choose. 
To use complimentary counting, consider all the ways Mary and Jane can sit together, so there are $2!(6!)$. The probability is thus $$\frac{7! - 6!(2!)}{7!} = \frac{5}{7}.$$
A: There are 6 places where they can sit together: (this is 7-2+1 for generalisation)

For each place, either John or Mary can sit on the left ($2$ ways) and each of the other 5 can sit in any order ($5!$ ways).
Thus there are $6*2*5! = 2*6!$ ways for John and Mary to sit together.
There are $7!$ permutations in total, thus the probability of sitting together is $\frac{2*6!}{7!} = \frac{2}{7}$.
Thus the probability of not sitting together is $1-\frac{2}{7} = \frac{5}{7}$.
A: Five seats have $2$ neighbour seats and two seats have $1$ neighbour seat. 
Hence each person sits with $\frac{5\cdot 2 + 2 \cdot 1}{7} = \frac{12}{7}$ persons on average. Now there are $6$ other persons than Mary, so the probability that she sits with John is $\frac{1}{6} \cdot \frac{12}{7} = \frac{2}{7}$.
Alternatively:
Add another seat to the arrangement making it a circle, and place the cinema owner in that seat. Then Mary and John sit next to eachother with probability $\frac{2}{7}$.
A: Out of $\frac{7 \times 6}{2}$ pairs of seats, 6 are adjacent giving a $1-\frac{2}{7}=\frac{5}{7}$ chance they sit apart.
A: Assume that Mary is the last to arrive.   There are 7 positions where she can insert herself into the row.  Two of those positions are next to John, no matter where he is sitting.   So if she chooses a position randomly, the probability she will sit next to John is 2/7.
A: 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

A: First seat the other 5 friends say K, C, G, D, and T without John and Mary.


*

*K * C * G * D * T *
Note that the stars represent places where John and Mary can possibly seat.
therefore, they can seat in 6P2 ways = 30 ways.
Then other friends can seat in 5! ways = 120 ways
Again consider seating them all together without restrictions for the purpose of probability space (7! = 5,040 ways) 
Take a product of the results of seating John and Mary and other five friends. Divide your answer with the result of seating all the 7 friends together without restrictions.
(30*120)/5040 = 5/7

