I am a beginner in Probability Theory. Upon reading and solving few problems from the first five chapters from Feller's - Introduction to Probability theory and applications book, I know that, every counting problem is abstractly equivalent to distributing $r$ balls to $n$ cells.
Consider the case $r=7, n=7$. The number of distinguishable distributions are $7^7$.
The number of distinguishable distributions with the occupancy numbers $2,2,1,1,1,0,0$ appearing in any order is given by:
(1) Label the balls.
$\begin{array}{ll} \text{Selection} & \text{Distinguishable choices}\\ \hline \text{Choose 2 balls to place in the first doubly occupied cell} & {7 \choose 2} \\ \text{Choose 2 balls to place in the second doubly occupied cell} & {5 \choose 2} \\ \text{Choose 1 ball to place in the first singly occupied cell} & {3 \choose 1} \\ \text{Choose 1 ball to place in the second singly occupied cell} & {2 \choose 1} \\ \text{Choose 1 ball to place in the third singly occupied cell} & {1 \choose 1} \\ \end{array}$
So, we've assigned $5$ labels to these balls (this is a $5$-tuple). This is just the multinomial coefficient $\frac{7!}{2!2!}$. One such choice is $\{ab,cd,e,f,g\}$. The order of the sub-populations is important, so $\{ab,cd,e,f,g\}$ is distinguished from $\{cd,ab,e,f,g\}$. No attention is paid to the order within the groups.
(2) Label the cells.
$\begin{array}{ll} \text{Selection} & \text{Distinguishable choices}\\ \hline \text{Choose 2 cells that will be doubly occupied} & {7 \choose 2} \\ \text{Choose 3 cells that will be singly occupied} & {5 \choose 3} \\ \text{Choose 2 empty cells} & {2 \choose 2} \\ \end{array}$
The number of distinguishable choices are $\frac{7!}{2!3!2!}$. This results in a $3$-tuple. One example would be $(12),(34),(567)$.
Now, when we multiply these quantities, to find the total number of distributions; it is simply like multiplying $2$ shirts with $3$ ties to get $2 \times 3 = 6$ shirt-tie pairs that can be worn on two consecutive days.
Question.
Since, $(12),(34),(567)$ is equivalent to $(21),(34),(567)$, my understanding is: in the above product, we don't distinguish between $\vert ab \vert cd \vert e \vert f \vert g \vert \vert \vert$ and $\vert cd \vert ab \vert e \vert f \vert g \vert \vert \vert$ - they count as a single choice. And it is so, because the occupancy numbers are the same; switching the two's in $2,2,1,1,1,0,0$ does not change the outer appearance of the configuration?