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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:

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

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    That's right. What is the question? – 2017-01-15
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    @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. :) – 2017-01-15

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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.