Birthday problem with exactly 3 people I have been doing a bunch of birthday problem questions, however this one has thrown a mental block my way. The questions is:
What is the probability that in a room of 10 people exactly 3 people will have the same birth month as each other, while the other 7 have different birth months from everyone else.
 A: The number of ways to choose $3$ out of $10$ people is $\binom{10}{3}$.
The number of ways to choose $8$ out of $12$ birth-months is $\binom{12}{8}$.
The number of ways to assign $8$ birth-months among them is $8!$.
The total number of ways for $10$ people to have birth-months is $12^{10}$.

So the probability is:
$$\frac{\binom{10}{3}\cdot\binom{12}{8}\cdot8!}{12^{10}}\approx3.868\%$$

Please note that by choosing $8$ birth-months, we are essentially choosing $1$ birth-month for the group of $3$ people, and $7$ different birth-months for each one of the remaining $7$ people.
A: the answer is $0.03868$ which is same as posted by Barak Manos.
firstly the problem assumes that all people in the room are born in the same year.
$10$ people can choose $12$ months in $12^{10}$ ways. This forms our sample space.---(1)
Now as the problem states that exactly $3$ people have their b'days in the same month, first we choose $3$ people from the $10$ people in $10C3$ ways. now as the $3$ people have their b'days in a particular month, that month can be selected in $12C1$ ways. for the remaining $7$ people there are $11$ months left. And since their b'days must not fall in the same month, the choosing of months is without replacement.
So the numerator becomes:  $10C3 * 12C1 *(11C1 * 10C1 * 9C1 * 8C1 * 7C1 * 6C1 *5C1)$....(2)
Note that the term in the bracket indicates the remaining $7$ people choosing their b'days.
So the required probability is (2) divided by (1)
