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.

Look i don’t know the answer to this but someone else will so just wait a little bit longer

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.