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?

• How so? ANd if that is the probability of them sitting together why did you multiply it by twelve? I don't follow your reasoning. You got that there are six places for the two to sit and two ways for the them to sit together but you haven't figured out how to use those numbers to get the probability. – fleablood Jun 22 '17 at 16:57
• That depends on whether any of the other five friends are named "John" or "Mary" ;) – Doktor J Jun 23 '17 at 16:36
• If they like each other, they'll see to it that they sit together. – René Nyffenegger Jun 24 '17 at 14:27
• @BenKovitz -- the problem says "randomly". – Malvolio Jun 25 '17 at 8:55
• @BenKovitz -- that's true, but you can drive yourself crazy trying to find loopholes in any word problem. It doesn't specify that each friend gets one seat, so if John ends up sitting in Mary's lap, does that count as "next to"? It doesn't specify that the seats are linear or continuous; perhaps there are gaps or the "row" is circular. A number of comments here are focused on whether "John" and "Mary" are unique identifiers of individuals. You have to look at the most reasonable interpretation of the problem. – Malvolio Jun 25 '17 at 16:04

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.

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.)

• You can easily generalise this to any $n > 1$ friends; the $6$ comes from $n - 1$ and $n\choose2$ computes to $\frac{n(n - 1)}2$ so the expression becomes $1 - \frac{2(n-1)}{n(n-1)}$ which simplifies to $1 - \frac{2}n$. – Neil Jun 23 '17 at 18:40
• @Neil, nice observation, thanks. The alternative solution also clearly generalizes, since no matter how many people there are, there will only be $2$ slots where Mary can slip in next to John. – Barry Cipra Jun 23 '17 at 20:41

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

• I'm afraid I don't understand your argument. I'd be most grateful if you could give me some more details. – Pierre-Yves Gaillard Jun 24 '17 at 15:24
• @Pierre-YvesGaillard: Barry Cipra in his alternative answer has given essentially the same argument in story form. Are you still having doubts ? – true blue anil Jun 24 '17 at 17:02
• Thanks a lot! I never doubted that your answer was correct, I just didn't understand it. Now that I do, I find it very nice! – Pierre-Yves Gaillard Jun 24 '17 at 17:09
• @Pierre-YvesGaillard: You're welcome ! – true blue anil Jun 24 '17 at 17:10

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

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

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 $2\cdot(\frac{1}{7}\cdot\frac{1}{6}) = \frac{2}{7}\cdot\frac{1}{6}$

• Seems like your $3/7$s in this answer should be $5/7$s. – Gregor Jun 23 '17 at 0:09
• This is probably correct, but you do not state clearly enough what you are doing in either your question or this answer. It seems that the answer should contain something like “I have to distinguish who of the two is sitting on the right either in all places of the solution or not at all” – Carsten S Jun 23 '17 at 9:45

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 $1-\frac{2}{7} = \frac{5}{7}$.

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 $\frac{5}{7}$

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!

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 $$\frac{7! - 6!(2!)}{7!} = \frac{5}{7}.$$

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 $\frac{2}{7}$.

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.

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.

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