# Confusion about counting in Feller's probability textbook

The problem is: Suppose there are 7 balls and 7 boxes. Given that two boxes are empty, how many ways are there to get a box with exactly 3 balls?

In Feller's Introduction to Probability Volume I, he claims that the number of ways to do so is $\frac{7!}{4!2!1!}\cdot \frac{7!}{3!1!1!0!0!}=88200$. This is because you want to find how many ways there are to get the partition 3,1,1,1,1,1 in some order. This can be thought of as consisting of two subpopulations: 4 boxes with one ball, 2 boxes with no balls, and 1 box with three balls. Hence the first term. The second term comes from permuting 7 balls in 7 boxes and dividing out the overcounting.

On the other hand, I find that there should be $7\cdot 6\cdot 5\cdot 4\cdot 5=4200$ ways. Consider the 5 boxes that are non-empty. There are 7 choices of balls for the first, 6 choices of balls for the second, 5 choices of balls for the third, 4 choices of balls for the fourth, and 1 choice for the last since you throw all 3 remaining balls into it. There are also 5 ways of choosing the box containing three balls.

The two approaches seem correct to me, but there of course can only be one answer. I feel like I must be making some hidden assumptions involving distinguishibility/ordering, but I can't figure it out. Can someone help explain?

Evidently, the balls and boxes are distinguishable. Suppose the seven boxes are lined up in some order. You must first choose which two of the seven boxes are to be left empty, which can be done in $\binom{7}{2}$ ways. That leaves you with five choices for which box is to receive three balls. We can select three of the seven balls to place in that box in $\binom{7}{3}$ ways. That leaves us with four balls to place in the remaining four boxes, which can be done in $4!$ ways. Thus, the number of ways to distribute the balls so that exactly two of the seven boxes are empty and exactly one box receives three balls is $$\binom{7}{2}\binom{5}{1}\binom{7}{3}4! = 88,200$$ Your calculation failed to take into account the $\binom{7}{2}$ choices of which two boxes would be empty. Also, $7 \cdot 6 \cdot 5 \cdot 4 \cdot 5 = 840 \cdot 5 = 4200$.
• Since the boxes are distinguishible, why don't we have to multiply by $88,200$ by $7!$? Commented Dec 21, 2016 at 2:11