Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the sample space 8 choose 2? Why? What would the answer to this be?

share|cite|improve this question
$\binom 8 2$ is a number, not a sample space. Moreover, where did 8 come from. I do not see it anywhere else in the question. – Barbara Osofsky Feb 6 '13 at 1:50
up vote 3 down vote accepted

Wherever Mike is seated, there are $7$ unoccupied seats, and Joe is equally likely to be at any of them. Of these $7$ seats, $2$ are next to Mike, so the probability Joe is next to Mike is $2/7$.

Remark: There can be more than one sample space that is useful for calculating a probability. If we are going to solve the problem by counting, the most important thing is to make sure that we use a sample space of equaly likely outcomes.

One possible sample space is the set of pairs of seats. That sample space has $\binom{8}{2}$ elements. Or else it may be useful to think of Mike as sitting down first, and then Joe. The set of ordered pairs of seat choices is then a good sample space. That sample space has $(8)(7)$ elements. Or else we can exploit the symmetry, and use as our sample space the set of choices Joe can make, after Mike is seated. That gives a nice small sample space with $7$ equally likely elements.

share|cite|improve this answer

HINT: If they’re seated in a circle, there are indeed $\binom82$ possible pairs of seats in which Mike and Joe might sit. There are $8$ pairs of adjacent seats, so the probability that they sit together is ... ?

share|cite|improve this answer
Sorry forgot to mention circular seating arrangement – John Feb 6 '13 at 1:42
@Koy: It makes only a small change: you get one extra pair of adjacent seats. – Brian M. Scott Feb 6 '13 at 1:44
Is it 8/28? But if we place Mike in one seat, aren't there 2 options of Joe? Wouldn't that make it 16? – John Feb 6 '13 at 1:53
@Koy: $\frac8{28}=\frac27$ is correct. You can also work it the way you’re thinking about, but then you have to realize that the denominator is not $\binom82$, but rather $8\cdot7$, because you’re counting ordered pairs of seats: there are $8$ ways for Mike to choose a seat, and once he’s chosen, there are $7$ ways for Joe to choose a seat, for a total of $56$ seat assignments, $16$ of which have them together. In other words, you can look either at how many of the $28$ possible pairs of seats are adjacent, or you can look at how many of the $56$ possible seatings of the two put ... – Brian M. Scott Feb 6 '13 at 1:56
... them next to each other. – Brian M. Scott Feb 6 '13 at 1:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.