Hey, I am trying to find the probability that exactly two people out of m have the same birthday. I know that the probability for at least two people out of m having the same birthday is, P(X>=2) = 1 - P(X!=2) = 1 - 365/365*364/365*363/365*...*(365-m+1)/365 = 365!/[(365-m)!*365^m] which is just 1 minus the probability that m people have different birthdays (P(X!=2)). Now suppose I want to find the probability that EXACTLY two (but no more) share a birthday, using similar logic, this is the probability that no more than two share the same bday, = 365/365*1/365*364/365*...*(365-m+2)/365 = 365!/[(365-m+1)!*365^m] but I think I am overlooking something. Does the case where, say, two students born on June 15th and two students born on July 3rd, with everyone else having different bdays, still fall under the case that EXACTLY two share a birthday (since there are no 3 students that share a bday?). If so, how do I do this problem?