The probability that exactly two people of N are born in a given month (if N<13, see end) is:

(1/12)^2 * (11/12)^(N-2) * N(N-1)/2

where the first factor is the two people being born in that given month, the second is all the rest not being born in that month, and the third is the number of ways that can happen: N people who could be one person born in that month, N-1 remaining people who could be the second person born in that month, and divide by 2 because we've counted each pair of people twice (there are 496 different pairs among 32 people, not 992).

For three people, it would be:

(1/12)^3 * (11/12)^(N-3) * N(N-1)(N-2)/6

where the final factor is N who could be the first, N-1 remaining who could be the second, N-2 remaining who could be the third, and divide by six because we've counted each trio six times.

In general, the odds that precisely X people of N were all born in a given month are:

(1/12)^X * (11/12)^(N-X) * N! / (X! * (N-X)!)

where the final factor has become the number of ways of choosing X items from N equivalent ones (a binomial coefficient).

For N=32, X=9, this evaluates to 0.000734, which multiplied by 12 (because there's nothing really special about August -- you'd be equally surprised if the nine shared any other birth month) is 0.0088, meaning that there's about a 1 in 113 chance that, of 32 randomly-selected people, exactly 9 will share a birth month.

Except that type of calculation is not exactly accurate -- for example, if N=13 or more then the chance that two or more people will share a birth month is 100%, by the pigeonhole principle. What I've missed out, is that the other people -- the 30 others in the N=32, X=2 case, or the 23 others in the N=32, X=9 case -- might also include X further people who share a birth month. In the X=2, N>=13 case, that's a certainty, so the result is dead wrong. But in the N=32, X=9 case, that's rather unlikely (maybe one chance in 2,000), so the answer I've given is, I think, close to correct.

Peter