Given $20$ people, what is the probability that among the $12$ months in the year, there are $4$ months containing exactly $2$ birthdays and $4$ containing exactly $3$ birthdays?

$20$ people are in the room:

Outcome space of the experiment is $S\left\{\left(x_1,x_2,x_3,\dots x_{20}\right):x_i\in \{1,2,3,\dots 12\}\right\}$ where the vector describes their months of birth.

$\left|S\right|={12}^{20}$ because each element of the $n$ valued vector can be chosen in 12 ways.

As each outcome from the outcome space is equally likely, the Axioms give:

$\left(\right|X|$ is the number of elements in the set $X$ )

Probability that the months of birth are grouped in $4$pairs and $4$triplets?

Let's calculate the number of sample events where this event happens.

Firstly permute $20$elements from which $4$pairs are equivalent, and $4$threes. This corresponds to every different order of months in the vector

$\left(\begin{array}{c}20\\ 2,2,2,2,3,3,3,3\end{array}\right)$

This already permuted the months with precisely two birthdays, and those with precisely three birthdays. Now the only thing left is to choose $8$out of $12$months $\left(\begin{array}{c}12\\ 8\end{array}\right)$ways, and count the partitions of those $8$months to the ones that have $3$birthdays, and those with two, and this is $\left(\begin{array}{l}8\\ 4\end{array}\right)$. This leaves us with:

localid="1650015410942" $$\begin{gathered} \left(\begin{array}{c}20\\ 2,2,2,2,3,3,3,3\end{array}\right)\times \left(\begin{array}{c}12\\ 8\end{array}\right)\times \left(\begin{array}{l}8\\ 4\end{array}\right)=\frac{20!}{2{!}^4·3{!}^4}\times \frac{12!}{8!4!}\frac{8!}{4!4!} \\ P\left(X\right)=\frac{\frac{20!·12!}{2{!}^4·3{!}^4·(4!{)}^3}}{{12}^{20}}=0.0010604\dots \end{gathered}$$