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?
Best Answer
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.