# 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?

• You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor. – Arturo Magidin Dec 17 '18 at 22:43
• @ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can. – Stan Shunpike Dec 17 '18 at 22:49

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.
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.