I want to know how many ways I can place some, possibly all of five balls each a distinct color into three distinguishable bins. Each bin must have at least one ball and I do not need to use all of the balls.
The numbers are small so I consider the cases:
(x,y,z) is the number of items in each bin. With only 5 items to use (at most) I have these possibilities:
(1,1,1) (5*4*3)/(1!1!1!)
(2,1,1) (5*4*3*2)/(2!1!1!)
(1,2,1) (5*4*3*2)/(2!1!1!)
(1,1,2) (5*4*3*2*1)/(2!1!1!)
(2,2,1) (5*4*3*2*1)/(2!2!1!)
(1,2,2) (5*4*3*2*1)/(2!2!1!)
(2,1,2) (5*4*3*2*1)/(2!2!1!)
(3,1,1) (5*4*3*2*1)/(3!1!1!)
(1,3,1) (5*4*3*2*1)/(3!1!1!)
(1,1,3) (5*4*3*2*1)/(3!1!1!)
Then take the total of the right column. I first count the ways of putting the balls in as if order mattered, then divide by the number of arrangements possible of balls that are grouped together in each of the three bins to account for repeats.
That's not a bad process for small numbers, but is there any way to streamline this to solve the problem?:
How many ways I can place some, possibly all of n balls each a distinct color in to m distinguishable bins? Each bin must have at least one ball and I do not need to use all of the balls.