5.16 How many ways are there to put 4 balls in 3 boxes if:I don't understand the part of the solution that explains how many ways you could put 2 balls in one box, 2 balls in another box, and 0 balls in a third box:
(b) the balls are distinguishable but the boxes are not.
(2,0,0): There are 4C2 = 6 ways to choose the balls for the first box, and the remaining go in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get 6/2 = 3 arrangements.To me, it seems like there would have to be 6 ways to choose 2 balls from balls 1, 2, 3, and 4 for the first box:
What am I missing?
For the option of (2,1,1), the solution is 6, not 3:
There are 4C2 = 6 options for picking the two balls to go in one box, and each of the other two balls goes into its own box.I don't see how (2,0,0) is different from (2,1,1) when the boxes are indistinguishable.