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