For permutations with “copies”, in what sense are the elements of the set distinct? Suppose I have a set of elements $$ S = \lbrace A,B,B,D,E \rbrace$$
Then the number of distinct permutations of size 5 is 
$$\frac{5!}{2!} = 60$$ 
This is because $S$ contains a ‘copy’ of B. 
But by definition, in a set of elements, each element is distinct. However, now I’m essentially saying “wait actually B and B aren’t distinct.”
So why isn’t the set $$ S = \lbrace A,B,D,E \rbrace$$ if the B elements are not distinct? Notice the permutations of this set are very different 
$$4! = 24$$ 
Are there two different kinds of being distinct here? 
 A: You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $\ldots B_1\ldots B_2\ldots$ have a seocnd equivalent one $\ldots B_2\ldots B_1\ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
A: take a simpler case $\{A,B,B\}$ vs $\{A,B\}$.  First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$.  Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
