Probability that among 5 people, exactly 2 of them are born in the same month 
Among 5 people, what is the probability that exactly 2 of them are born in the same month. 

For "at least $2$ in the same month", my answer would be 
$$1 - \frac{12\cdot 11\cdot 10\cdot 9\cdot 8}{12^5}$$
the complement of probability that all months are distinct. How to deal with "exactly 2"?
(Assuming the birthdays are independent, and equally distributed among the months.) 
 A: We'll consider 12 different cases: 
(1) Exactly two people are born in Jan and the remaining three are born in different months.
(2) Exactly two people are born in Feb and the remaining three are born in different months.
... and so on. Notice that the cases are mutually excluding, i.e., an event were exactly two people are born the same month can't be in two cases at the same time. Hence, the sought probability would be the sum of probabilities of each case.
What's the probability that two people are born in Jan and the rest during different months? There are ${5\choose2}$ ways to pick 2 people out of 5. There is a $\frac1{12}$ chance that a person is born in Jan (since we're assuming births are uniformly distributed), so there is a $(\frac1{12})^2$ that two people are born the same month. In this case, is Jan. Now we want the third person to be born in any month other than Jan. There are 11 months to choose from. Once chosen, the fouth person must be born in a different month than Jan and the one where the third person was born. Thus, there are 10 months to pick from. And so on. 
Hence, it would be a $p={5\choose2}(\frac1{12})^2\cdot \frac{11}{12}\cdot\frac{10}{12}\cdot\frac9{12}$ chance that exactly two people are born in Jan and the remaining three are born in different months.
Since the months and births are independent from each other, the 11 remaining cases are exactly the same.
Then your answer would be $12p$
A: It is tacitly assumed that there are $12^5$ equiprobable cases. We now have to count the favorable cases. There are ${5\choose2}=10$ ways to select the two people having birthday in the same month, and $12$ ways to select that month. There are three people left, to which we can assign a month each in $11\cdot10\cdot 9$ ways. The probability in question therefore comes to
$$p={10\cdot 12\cdot 11\cdot 10\cdot 9\over 12^5}={275\over576}\doteq0.47743\ .$$
A: Hint : Fix two persons and a month. ($120$ possibilities so far). Now, choose three other months ($165$ possibilities). In each case, you have $6$ possibilities to distribute the three months over the three remaining persons. 
Do not bother Ittay's comment. Even, if the probabilities are not equal, it is OK to ask for the probability assuming equality.
