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?

Solution 1:

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.

Solution 2:

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.)

Solution 3:

$\dfrac{6\cdot5}{6\cdot7} = \dfrac57\quad$ 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 $Pr = \dfrac{6\cdot5}{6\cdot7} = \dfrac57$

Solution 4:

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 $$1 -\frac{2 \cdot 6 \cdot 5!}{7!} = 1 -\frac{2 \cdot 6!}{7!} = 1 - \frac{2}{7} = \boxed{\frac{5}{7}}$$