Probability of at least two persons in four share birth-months, without complement probability Assume we have four people, how to calculate probability of at least two of them share birth-months?

Using complement probability, we know it will be
$ P(A) = 1 - P(A') = 1- \frac{12\times11\times10\times9}{12^4} $, but how could it be calculated without complement?
 A: There are $12^4 = 20736$ possible ordered sequences of birth months.
According to your computation, 11,880 have no birthday matches.
That leaves 8856 with at most one match.
In order to solve the problem without using complementation, you would have to look at all of the ways to get one or more matches
among birth months. (Cheating above, we know the sum of them must
be 8866.)
There are 12 of these in which all 4 people share a birth month.
Also, there are 6 * 12 * 11 * 10 = 7920 ways to have exactly one birth month match. That is, two people born in the same month
and the other two born in two other different months. The 6 enumerates placements of the two matching birth months in the
sequence of 4.
Then you have to figure out the number of ways for '3 and 1'
and for '2 and 2' (both of which result in two redundant birth
months among the four).
A simulation with a million performances of the experiment very roughly verifies some of the above. (Of course we know from above that
the first number should be 11880, the second should be 9720, and 
that the last number is exactly correct. The third number is the
sum of the numbers of ways for the two configurations mentioned
above.)
 m = 10^6;  x = numeric(m)
 for (i in 1:m) {
   b = sample(1:12, 4, rep=T); x[i] = length(unique(b)) }
round(12^4*table(4-x)/m)

##     0     1     2     3 
## 11894  7910   920    12    # rough values

Note:  So it can be done this way, but why would you want to do
so, when complementation is easier?
A: What you are asking for is $$\frac{\binom{12}{1}\binom{4}{2}\binom{11}{2}\binom{2}{1}\binom{1}{1}+\binom{12}{2}\binom{4}{2}\binom{2}{2}+\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{1}{1}+\binom{12}{1}\binom{4}{4}}{12^4}=\frac{8856}{12^4}$$
The terms on top are counting the various ways in which you could have two people sharing a birth month. For instance, $\binom{12}{1}\binom{4}{2}\binom{11}{2}\binom{2}{1}\binom{1}{1}$ counts when two share a birth month and the other two have two different birth months. The product represents choosing a birth month for two people to share, then choosing which two share it, then choosing two months for the other two people, then choosing which of those two people gets the earlier month, and then as a formality choosing which of the last one person gets the last chosen month.
The next term considers two people who share a birth month and another two people who share a different birth month. 
The third term considers three people who share a birth month, with the fourth person having a different birth month.
And the last term considers when all four have the same birth month.
