I have the following exercise, how many ways are there two rearrange the letters in the word “apple”. I know that the answer will be $5!/2$, as we have $5!$ permutations, but as we have $2$ "p"s we counted twice every different permutation.
I would like to get the same result using a more formal argument. We had in the lecture the following argument (which is I think used implicitly in my argumentation):
Let $Y,X$ be sets. Suppose we have a disjoint partition $ Y = \dot\bigcup_{x\in X} Y_x$ with $\#Y_x =m$. Then $\#X = \frac{\#Y}{m}$. Am I right that this is used implicitly and what are my $Y_x$ precisely.
To answer your question, yes, the theorem you mentioned is being used implicitly. Here are the details of how that theorem is applied to "APPLE" permutation problem.
$Y$ is the set of permutations of the word $AP_1P_2LE$. All letters are distinct, since $P_1$ and $P_2$ have been given labels that distinguish them. Therefore, $\#Y=5!$.
$X$ is the set of permutations of the word $APPLE$, where the two copies of $P$ are identical.
For each $x\in X$, we let $Y_x$ be the two resulting words that result from labeling the $P$'s in $x$ in two different ways. For example, if $x=PAPEL$, then $Y_x=\{P_1AP_2EL,\;P_2AP_1EL\}$. Note that each $\#Y_x=2!$, because there are $2$ copies of the letter $P$ in each rearrangement, which can be labeled with $1$ and $2$ in $2!$ ways.
Since $Y$ is the disjoint union of $Y_x$ over all possible $x\in X$, we conclude that $\#X=\#Y/\#Y_x=5!/2!$.