Probability of birthdays of thirty people Given thirty people, how to find the probability that among twelve months there are six months containing two birthdays and another six containing three birthdays.
Solution:- Solution given is $\left(\frac{30!}{2^6*6^6}\binom{12}{6}\right)*12^{-30}$
I agree that total sample space is $12^{30}$ and any six months can be selected out of twelve months in $\binom{12}{6}$ ways. I didn't understand this figure$\frac{30!}{(2^6)*(6^6)}$. What is the author's logical thinking behind it?
 A: First find the probability that there are $2$ birthdays in each of the first $6$ months and there are $3$ birthdays in each of the other months. 
The probability on that event is:$$\frac{30!}{2!^63!^6}\left(\frac1{12}\right)^2\cdots\left(\frac1{12}\right)^2\left(\frac1{12}\right)^3\cdots\left(\frac1{12}\right)^3=\frac{30!}{2!^63!^6}12^{-30}$$
This is multinomial distribution (actual $12$-nomial distribution with $12$ equal parameters).
Now realize that there are $\binom{12}6$ different choices for the $6$ months having $2$ birthdays. 
This leads to probability: $$\binom{12}6\frac{30!}{2!^63!^6}12^{-30}$$
A: Here's one way to get that figure.
To identify one of the ways the $30$ people can have birthdays satisfying the question's conditions, out of the $12^{30}$ possible ways the $30$ people can have their birthdays distributed among the $12$ months,
it is not enough to identify which of the months have two birthdays and which have three.
It also matters which two people have birthdays in the first two-birthday month, which have birthdays in the second, and so forth.
If we were to enumerate all those possible combinations, we would have
$\binom{30}{2}$ ways to choose who is born in the first two-birthday month.
That leaves $28$ people yet to be assigned, and $\binom{28}{2}$ ways to choose who is born in the second two-birthday month.
Note that
$$\binom{30}{2} \binom{28}{2} = \frac{30!}{2!28!}  \frac{28!}{2!26!} 
= \frac{30!}{2! 2! 26!}.$$
So we get some cancellation when we expand the binomial coefficients to factorials.
The next factor, $\binom{26}{2},$ will cancel the $26!$ in the denominator, and so forth, and in the end (after running through all the two-birthday months and three-birthday months) you have
$$ \frac{30!}{2! 2! 2! 2! 2! 2! 3! 3! 3! 3! 3! 3!} = \frac{30!}{2^6 6^6}.$$
Alternatively, using multinomials, the number of ways to distribute $30$ distinguishable people into a sequence of six groups of two and six groups of three
(the groups all having unique identities, in this case month names, but the sequence of individuals within a group not being considered), is
$$ \binom{30}{2\ 2\ 2\ 2\ 2\ 2\ 3\ 3\ 3\ 3\ 3\ 3}
=\frac{30!}{2! 2! 2! 2! 2! 2! 3! 3! 3! 3! 3! 3!},$$
same as before.
(In fact I would take the previous argument as an intuition for the multinomial formula, though further proof is required to make the formula general.)

A slightly faster way to get the result if you're not already familiar with multinomials:
to distribute the $30$ birthdays among the twelve months,
we line the $30$ people up in a row according to the order of their birthdays.
This gives $30!$ ways to distribute the birthdays, but since we did not respect the order of birth within the month when we came up with the $12^{30}$-element sample space,
we have overcounted each actual element of the sample space by a factor of $2!$ for each two-birthday month (since we separately counted both ways those birthdays could occur) and a factor of $3!$ for each three-birthday month,
so we have to divide by $2!$ six times and by $3!$ six times to get back to the correct number.
This is similar to an argument you could have made for "$n$ choose $k$" formula,
except that we have more than two groups into which we are partitioning the $n$ items.
