# Feller's Introduction to probability - Configuration of 7 balls in 7 cells

I am not sure, if my understanding of the below problem is correct. Any inputs/insight would be extremely helpful.

On page 38, Feller writes :

Configuration of $$r=7$$ balls in $$n=7$$ cells.

For the sake of definiteness, let us consider the distributions with occupancy numbers $$2,2,1,1,1,0,0$$ appearing in an arbitrary order. These seven occupancy numbers induce a partition of seven cells into three sub-populations (categories) consisting respectively, of the two doubly occupied, the three simply occupied and the two empty cells. Such a partition into three groups can be effected in $$7!/(2!3!2!)$$ ways.

To each particular assignment of our occupancy numbers to the seven cells, there correspond $$7!/(2!2!1!1!0!0!)$$ different distributions of the $$r=7$$ balls into the seven cells.

Accordingly, the total number of distributions such that occupancy numbers co-incide with $$2,2,1,1,1,0,0$$ in some order is:

$$\frac{7!}{2!3!2!}\times\frac{7!}{2!2!1!1!0!0!}$$

First term.

I understand that the first term on the left, are the possible permutations of two doubly-occupied, three singly occupied and two empty-cells, that is, orderings of the occupancy numbers. For example, $$\{2,2,1,1,1,0,0\},\{2,1,2,1,1,0,0\},\ldots$$. We are just partitioning $$n=7$$ cells into $$3$$ sub-populations.

Second term.

Suppose we have, $$7$$ balls labelled $$1,2,3,4,5,6,7$$. We are to classify them into groups of size $$2,2,1,1,0,0$$. This gives $$7!/(2!2!1!1!1!0!0!)$$ possible groupings.

$$\implies$$ the possible distributions with occupancy numbers $$2,2,1,1,0,0$$ in an arbitrary order is:

$$\frac{7!}{2!3!2!}\times\frac{7!}{2!2!1!1!1!0!0!}$$

• That's right. What is the question? Commented Jan 15, 2017 at 3:39
• @leonbloy, how are we making sure, the balls in pairs don't go to a cell with occupancy number $1$. The explanation above hasn't entirely sunk in. :) Commented Jan 15, 2017 at 3:47

It might be more helpful to view the second term as $$\binom{7}{2} \binom{5}{2} \binom{3}{1} \binom{2}{1} \binom{1}{1},$$ as there are $\binom{7}{2}$ ways to select 2 balls to place in the first group of size 2, $\binom{5}{2}$ ways to select 2 of the remaining balls to place in the second group of size 2, etc.