student birth month distribution probability question

[Proggy99]

1. The problem statement, all variables and given/known data
A class contains 30 students. What is th eprobability that there are six months each containing the birthdays of two students, and six months each containing the birthdays of three students. Assume that all months have the same probability of including the birthday of a randomly selected person.

2. Relevant equations
distributing n students across k months with n1 students in the first month, n2 students in the second, ..., nk students in the kth month would use the formula n! / (n1!n2!n3!...nk!)
there are 12! variations on the above formula
there are 12^{30} possible combinations of 30 students spread across 12 months


3. The attempt at a solution
[(n! / (n1!n2!n3!...nk!)) * 12!] / 12^{30} =
[(30!/(2!2!2!2!2!2!3!3!3!3!3!3!)) * 12!] / 12^{30} = 179.255

I know the answer is .000346, I just do not know what I am doing wrong.

[Dick]

1. The problem statement, all variables and given/known data
A class contains 30 students. What is th eprobability that there are six months each containing the birthdays of two students, and six months each containing the birthdays of three students. Assume that all months have the same probability of including the birthday of a randomly selected person.

2. Relevant equations
distributing n students across k months with n1 students in the first month, n2 students in the second, ..., nk students in the kth month would use the formula n! / (n1!n2!n3!...nk!)
there are 12! variations on the above formula
there are 12^{30} possible combinations of 30 students spread across 12 months


3. The attempt at a solution
[(n! / (n1!n2!n3!...nk!)) * 12!] / 12^{30} =
[(30!/(2!2!2!2!2!2!3!3!3!3!3!3!)) * 12!] / 12^{30} = 179.255

I know the answer is .000346, I just do not know what I am doing wrong.

There aren't "12! variations on the above formula". The number of variations is the number of ways to choose six months out of twelve.

[Proggy99]

There aren't "12! variations on the above formula". The number of variations is the number of ways to choose six months out of twelve.

ah yes, that makes sense, I was focusing in the wrong direction when trying to figure out what I was missing. So by figuring out all the different variations of six months, I automatically have all the variations of the other six months in relation to the first six. I was doing 12! to do all the different orders of 12 months, which in this case does not fit the problem. Thanks for the help