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10 Students are randomly seated in a row for a photograph. Three of these are very close friends.

What is the probability that exactly two of the three friends will sit next to each other?

The solution that is posted is, The number of ways to choose two friends from the three friends is 3C2. The number of outcomes for which the two chosen friends sit next to each other at one of the ends of the row and the third friend has a non-adjacent seat is 2 x 7 x 2! X 7!, while the number of outcomes for which the two chosen friends sit next to each other but not at one of the ends of the row and the third friend has a non-adjacent seat is 7 x 6 x 2! x 7!. Hence the probability that exactly two of the three friends will sit next to each other is: 3C2 ((2x7x2!x7!) + (7x6x2!x7!)) / 10!

how to get (2x7x2!x7!) + (7x6x2!x7!) ??

  • 3
    And how many students are there initially? – 2017-02-23
  • 1
    Assuming there are $n$ seats available in the row, how many ways can you choose three of the $n$ seats to be used by the friends. Are each selection of three seats actually going to be equally likely to be the three seats used by the friends? Now... how many ways can you pick the three seats such that two of them are adjacent and the other is not? Suggest breaking into steps by picking the pair of seats first and the third seat after noting that the number of choices changes based on whether the pair was on the edge or not. – 2017-02-23

1 Answers 1

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Bind 2 together, turn them into a block. Then distribute the rest.

There $3 \cdot 2$ ways any of the 2 of the 3 friends may be together.

there are $(n-1)!$ ways to distribute the students if $2$ are a block.

there are $n!$ ways to distribute the students in total.

$\frac {3 \cdot 2 (n-1)!}{n!}$

but we have over-counted the possibility that all three are together.

there are $3!(n-2)!$ ways to distribute students keeping all 3 together.

$\frac {3! ((n-1)! - (n-2)!}{n!} = \frac {6(n-2)}{n(n-1)}$