How many ways can a group of 12 fraternity brothers be assigned to rooms a, b, and c, with 4 assigned to room a, 6 to room b, and 2 to room c?

Assign 4 brothers to room A in C (12,4) = 12! / (4!8!) ways

Assign 6 brothers to room B in C (8,6) = 8! / (6!2!) ways

Assign 2 brothers to room C in C (2,2) = 2! (2!0!) ways

Total number of ways is the product

C (12,4) * C (8,6) * C (2,2) = 12!8!2! / (4!8!6!2!2!0!) = 12! / (4!6!2!) = 13860 ways

Incidentally, this is the same answer as the number of permutations of arranging 12 objects in a line, with 6,4 and 2 objects being identical

P (4,6,2) = 12! / (4!6!2!) = 13860.

Think about how this relates to the problems of the brothers.