How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?
Assuming that we can have empty boxes....let k be the number of balls and n be the number of boxes
The number of ways =
C ( k + n - 1 , n - 1 ) =
C ( 6 + 3 - 1 , 3 - 1) =
C ( 8, 2) =
28 ways
Just use stars and bars 8C2 is 8!/6! * 2 which is equal to 28
Ok, Brainless, the last time I checked 8!/6! * 2 is not equal to 28 nor 8C2
I think brian meant
8!
____ = 28
6! 2!
