Easy problem (hopefully).
How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?
Thanks in advance!
Assuming that we can have empty boxes
Let k = the number of balls
Let n = the number of boxes
The total ways =
C ( k + n - 1 , n - 1 ) =
C ( 6 + 3 - 1 , 3 - 1 ) =
C ( 8 , 2 ) =
28 ways
