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

I am less concerned about knowing the answer than making sure I have thought about the problem correctly. 
The general problem seems to be finding the distinct integer-valued vectors that satisfy
$$x_1 + x_2 + ... x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, ..., 12$$
It can be proven that there are $\binom{20+12-1}{12-1}$ such vectors. I am presuming this to be the sample space.  (Please correct me if I'm wrong)
$4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20$, so there are $3$ distinct groups of $4$ months. Therefore we have $\binom{12}{4\ 4\ 4}$ possible ways to get the desired distribution of months. 
The probability I get is $$\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} $$. The answer in the back of the textbook is $1.0604 \times 10^{-3}$ so my answer is off. I just want to know where I went wrong. 
 A: Your count of the sample space treats the $20$ people as indistinguishable, when in fact they are not. The actual size of the sample space is $12^{20}$: each of the $20$ people can have a birthday in any of $12$ months. (We are making the simplifying assumption that the probability of being born in any given month is $\frac1{12}$; this is in fact false.)
To get the correct numerator, observe (as in effect you did) that there are $\binom{12}4$ ways to choose the $4$ months containing exactly $2$ birthdays and then $\binom84$ ways to choose the $4$ months containing exactly $3$ birthdays, so there are $\binom{12}4\binom84=\binom{12}{4\;4\;4}$ ways to choose the months. Then, however, the $20$ people have to be assigned to these months, and there are $$\binom{20}{3\;3\;3\;3\;2\;2\;2\;2}=\frac{20!}{12^4}$$ ways to do that. The desired probability is therefore
$$\frac{\binom{12}8\binom8420!}{12^{24}}\approx0.00106042\;.$$
A: I was comparing answer by Brian M Scott to this question to this answer of another similar question (lets call it 2nd question and this problem as $1$ st problem). I had following doubt:
$2$ nd problem, case $i=2$ solution has $\binom{4}{1,3}+\binom{4}{2,2}+\binom{4}{3,1}$ in the numerator.
$2$ nd problem, case $i=3$ solution has $\binom{4}{1,1,2}+\binom{4}{1,2,1}+\binom{4}{2,1,1}$ in the numerator. Then why Brian's solution dont have sum of different multinomial coefficients in its numerator? For examles, why there is no $\binom{20}{3,3,3,3,2,2,2,2}+\binom{20}{3,3,3,2,3,2,2,2}+\binom{20}{3,3,2,2,3,3,2,2}+...$ in the numerator of Brian's solution for problem 1?
I gave it a thought for a while. I felt that in 2nd problem we are selecting repairmen which can be permuted. That is different repairman can be assigned different number of TV sets. Thus this is imitated combinatorically by having sum of different multinomial coefficients. In 1st problem we dont select individual months. But select two groups each of 4 months, each month in one group getting 2 birthdays and in other getting 3. Selecting different months in different groups serves as a permuting 2 or 3 birthdays to these different months. Hence we dont have to consider different multinomial coefficients as all months in one group get same number of birthdays.
So I started thinking whether I can convert answer of 1st question to have sum of multinomial coefficients in the numerator. I felt, for this I need to select individual months. We can select $8$ out of $12$ months in $\binom{12}{8}$ ways. We can distributed 20 birthdays in four groups of 3 and four groups of 2 in  $\binom{20}{3,3,3,3,2,2,2,2}\times \binom{8}{4}$ ways, where $\binom{8}{4}$ counts different sequences of $\{3,3,3,3,2,2,2,2\}$ we can have in first multinomial's $2$nd parameter. So the final solution is $$\frac{\binom{12}{8}\times \binom{20}{3,3,3,3,2,2,2,2}\times \binom{8}{4}}{12^{20}}$$ which turns out to be same as Brian's answer, that is approximately $0.00106042$
