I have another good probability question that I can't answer.

An ordinary six-sided die is rolled 12 times. If X1 is the number of 1's and X2 is the number of 2's and so on, give the probability for:

a. P(X1=2, X2=3, X3=1, X4=0, X5=4, X6=2)

b. P(X1=X2=X3=X4=X5=X6)

Never mind, I figured it out.

The answer is

12!/(2!3!4!2!) * (1/6)^12


Thanks anyway!