Probability of birthday arrangements 
What is the probability that:

*

*the birthdays of 12 people will fall in 12 different calendar months (assume equal probabilities for 12 months);


*the birthdays of 6 people will fall in exactly two calendar months.

 A: The probability of the second person having birthday on a different calendar month of the first person is $\frac{11}{12}$, the probability of the third person having a different calendar month birthday than the first two people is $\frac{10}{12}$, so on. Therefore the probability of 12 people having their birthdays on different calendar months is $\frac{11}{12}\times\frac{10}{12}\times\frac9{12}\times\cdots\times\frac3{12}\times\frac2{12}\times\frac1{12}=\frac{11!}{12^{11}}$.
The probability that the third person having birthday on a month either of the first person's or the second person's is $\frac2{12}$, so is the fourth, fifth and sixth since only birthdays on two months are allowed.  Therefore the probability that the birthdays of $6$ people fall exactly in two months is $(\frac16)^4$.
A: While the first answer of user67258 is correct, the second is not. If the six people have birthdays in exactly two different months (not more, not less), then there are ${12 \choose 2}{(2^6 - 2)}$ equally likely ways to achieve this. The probability of the six people having birthdays in exactly two months thus equals:
$$\frac{{12 \choose 2}{(2^6 - 2)}}{12^6} = \frac{66 \cdot 62}{12^6} = \frac{4092}{2985984} \approx 0.00137$$
