probability 5 people have birthdays in 2 months The Question: What is the probability that in a family of 5, all birthdays will be in just two different months? 
What is wrong with the logic (12*11*2*2*2)/(12^5) (12 choices for the first person, 11 choices for the second person because they have to be different, and then the rest of the birthdays can be one of 2 months?)
 A: Well it doesn't matter what order you pick them in, which your working implies.
Hmmm this is rough working out and may be completely wrong, so be forewarned.
Choose the 2 months $\binom{12}{2}=66$ ways.
Now every person can be in either of the 2 months. As there are 5 people there are $2^5=32$ possibilities. We then subtract 2 cases (where their birthdays are all in one month or the other). This leaves us with 30 possible cases.
As you pointed out, every person can be in any of the 12 months, so that means there are in total $12^5$ possibilities.
$\frac{66*30}{12^5}=\frac{55}{6912}$ is what I get. 
Note that the problem is a bit flawed since both the answer and the question is based on the premise that the probability of a birthday being in a given month is $\frac{1}{12}$. 
A: There are two possible interpretations of the question: (1) what is the probability that all birthdays are concentrated in just two months, (2) what is the probability that all birthdays take place in exactly two months. You're trying to answer (2).
Your reasoning assumes that person 1 and person 2 have a birthday in two different months, but that is not necessarily the case. Instead, you should go over all possible patterns. Suppose person 1 is born in month A. For the rest of the people, there are $2^4-1$ possibilities, excepting the case in which all of them were also born in month A. There are $12\cdot 11$ possibilities to choose month A and month B (the other month), resulting in a probability of $12\cdot 11 \cdot (2^4-1)/12^5$.
For answer (1), you need to add $12/12^5$, which is the probability of the disjoint event that all birthdays were in the same month.
A: I'm going to start with the simpler "all family members have birthdays that fall within exactly two months, not just one" variant.  That logical predicate is satisfied if there is a person $n$ in the order such that:


*

*Anyone between person 1 and person $n$ in the order shares person 1's month ($\frac{1}{12}$ probability per person);

*Person $n$ does not share person 1's month ($\frac{11}{12}$ probability);

*Anyone after person $n$ in the order shares either person 1 or person $n$'s month ($\frac{2}{12}$ probability per person).


With 2 and 5 being the highest and lowest possible values for $n$, the probability of meeting the predicate for a given $n$ is:
$\frac{1}{12}^{n-2} \cdot \frac{11}{12} \cdot \frac{2}{12}^{5-n}$
Adding together those probabilities for $n$ where $2 \le n \le 5$ gives us the chance that the predicate will be met for any of those values of $n$ - which is the answer to our original problem.  If we want to solve the variant where it's acceptable for all five family members to share one month, we just add the probability of that event, which is $\frac{1}{12}^4$.
(Credit where credit's due:  I couldn't have worked out the answer without seeing Yuval Filmis's answer.  After a certain point, however, I couldn't follow him, so I thought I might well be able to give a clearer answer.)
