I have the following problem:

Determine the probability that out of 5 people sitting on a bench, 3 of them are sitting next to each other.

I know this is a simple problem, but I'm unsure my logic is correct.

My approach:

I first calculate the ways those 3 people are sitting next to each other - $3!$ and then I consider this as a one person so there are 3 ways this group of 3 people sit anywhere on the bench among the other 2. I divide by the total number of ways.


Is this correct?

Thanks in advance!

  • 2
    $\begingroup$ Also the 2 "outsiders" can sit in $2!=2$ different ways. So you must multiply by $2$. $\endgroup$ – drhab Aug 28 '17 at 15:15
  • 2
    $\begingroup$ Good start. But take the block $ABC$, suppose $A$ sits in the first seat. Then there are two possibilities. Namely $ABCDE$ and $ABCED$. $\endgroup$ – lulu Aug 28 '17 at 15:15
  • $\begingroup$ There are actually $3!$ ways for the block of three people to sit anywhere on the bench among the other $2$ since we have three objects to arrange, the block and the other two people. Observe that $$P(A) = \frac{3!3!}{5!} = \frac{3}{10}$$ $\endgroup$ – N. F. Taussig Aug 28 '17 at 16:40

Your approach is partially correct, but you also need to consider that the people outside the group may stay in any order. So, completing your proof: $$P(A) = \frac{18}{5!} \cdot 2 = \frac{36}{120} = \frac{3}{10}$$

As an example, if the three people are X, Y and Z, the people may be in any of the orders below (for a certain position, i.e. __XYZ):




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