Can one produce permutations from permutations on the same elements? I've just started studying combinatorics and I'm currently trying to understand permutations. So far I'd like to say that I do understand:

*

*Permutations without repetitions

*Partial permutations without repetitions

*Permutations with repetition

My question is if one can produce permutations out of permutations on the same elements?
Say for the 5 elements:
(A B C C D)
I first produce the possible permutations  with only 2 of these 5 elements, $$5!/3!=20$$ permutations.
Now out of the 20 permutations with only 2 elements, I want to produce the permutations with regards to repetition, or in other words, filter out the repetitions that occur among the 20 permutations. Doing it manually by hand I came to the conclusion that there are 7 repetitions among the 20, so the amount of permutations I am looking for are 13.
Is there a way of producing these 13 permutations out of formulas? I'm not necessarily asking for the solution (unless they only involve the 3 types of permutations I mentioned), but if it is possible and which method(s) I should read about concerning such a problem. I found something called "Permutations on multisets", but I'm not really sure that's it?
 A: Consider cases:

*

*A permutation of two distinct letters drawn from the multiset $\{A, B, C, C, D\} = \{A, B, 2 \cdot C, D\}$

*A permutation of two indistinguishable letters drawn from the multiset $\{A, B, 2 \cdot C, D\}$
A permutation of two distinct letters drawn from the multiset
$\{A, B, 2 \cdot C, D\}$:  There are four distinct letters in the multiset.  Therefore, there are four choices for the first element in the permutation and three choices for the second element since it must differ from the first element.  Hence, there are $4 \cdot 3 = 12$ $2$-permutations of distinct letters of the multiset $\{A, B, 2 \cdot C, D\}$
A permutation of two indistinguishable letters drawn from the multiset $\{A, B, 2 \cdot C, D\}$:  The only way to draw two indistinguishable letters from the muliset $\{A, B, 2 \cdot C, D\}$ is to draw both $C$s, which can be done in one way.
Since these two cases are mutually exclusive and exhaustive, the number of distinguishable $2$-permutations of the multiset $\{A, B, 2 \cdot C, D\}$ is $12 + 1 = 13$.
A: First of all, for this particular problem, I prefer N.F.Taussig's direct approach to the one that I will present in this answer (i.e. an indirect approach).  However, for other similar problems, sometimes an indirect approach will be easier.
Alternative approach:
You start with the computation of
$$\frac{5!}{(5-2)!} = 20.$$
Now, you look for a way of enumerating the repetitions from this group, so that they can be deducted.
There are two mutually exclusive ways that a repetition can occur:

*

*If the two items selected contain exactly one C.

*If the two items selected contain both C's.

To facilitate the enumeration, assume that the (C) elements are labeled C-1 and C-2.
There are $3 \times 2 = 6$ repeated items of form
(C-1,x) or (x,C-1) $~: ~\text{x} \in \{\text{A}, \text{B}, \text{D}\}.$
The repetition comes from (C-2,x), (x,C-2).
There is also one more repetition.
In the $5 \times 4$ computation, the permutations (C-1,C-2) and (C-2,C-1) are also counted separately.  So, to remove repetitions, one of these permutations must also be deducted.
Therefore, the final enumeration is
$$\frac{5!}{(5-2)!} - \left[3 \times 2\right] - 1.$$
