So, I have this task:
Find the probability that for 30 people between 12 months of the year, 6 months contain two, and the other six months for three of their birthdays.
and the solution given is $$\frac{30!}{2^6 6^6}\pmatrix{12\\6}12^{-30}.$$
Can someone explain the solution? Thanks.
The answer for arranging the months with two or three birthdays comes from the multinomial distribution: $$P(2,2,2,2,2,2,3,3,3,3,3,3)$$ $$=\frac{30!}{(2!)^6(3!)^6}\frac1{12^{30}} = \frac{30!}{2^66^6} \frac{1}{12^{30}}$$
Now, we can select the six months in $\binom{12}{6}$ ways where each month has the birthdays of two people each.
This gives us the required probability by just multiplying the two terms.