What is the correct terminology for Permutation & Combination formulae that allow repeating elements. Let me explain by example.
Q: Given four possible values, {1,2,3,4}, how many 2 value permutations are there ?
In "permutations", the order matters. In "combinations", the order does not matter.
The basic rules of counting are the Product Rule and the Sum Rule. See here, for example.


*

*Permutations with repetitions allowed:
If you have $n$ objects, and you want to count how many permutations of length $m$ there are: there are $n$ possibilities for the first term, $n$ for the second term, $n$ for the third term, etc. So the total number is $n^m$.

*Permutations without repetitions allowed:
If you have $n$ objects, and you want to count permutations of length $m$ with no repetitions (sometimes called "no replacement"): there are $n$ possibilities for the first term, $n-1$ for the second (you've used up one), $n-2$ for the third, etc. So the total number is
$$P^n_m = n(n-1)(n-2)\cdots(n-m+1) = \frac{n!}{(n-m)!}.$$

*Combinations without repetitions allowed:
In combinations we don't care about the order. If you select $m$ objects from $n$ possibilities and they are all distinct, then there are $P^m_m = n!$ ways of ordering them. So if you count permutations instead, you will count each distinct choice of elements exactly $m!$ times. Therefore, $\binom{n}{m}$ (also denoted $nCm$, both read "$n$ choose $m$") is given by:
$$\binom{n}{m} = \frac{1}{m!}P^n_m = \frac{n!}{m!(n-m)!}.$$

*Combinations with repetitions allowed:
You have $n$ objects, and you want to select $m$ of them, allowing repetitions. To simplify things, let us assume our $n$ objects are precisely the numbers $1$, $2,\ldots,n$. Given a selection of $m$ objects with possible repetitions, $a_1,\ldots,a_m$, order them in ascending order, $a_1\leq a_2\leq\cdots\leq a_m$, and think of them as an $m$-tuple: $(a_1,a_2,\ldots,a_m)$. We want to count the number of such tuples with entries between $1$ and $n$, repetitions allowed.
Let us associate to each such tuple an $m$-tuple with no repetitions and with entries between $1$ and $n+m-1$ as follows: given $(a_1,a_2,\ldots,a_m)$, with $1\leq a_1\leq a_2\leq\cdots\leq a_m\leq n$, we associate the tuple $(a_1,a_2+1,a_3+2,\ldots,a_m+m-1)$. Note that 
$$1\leq a_1\lt a_2+1\lt\cdots \lt a_m+m-1\leq n+m-1.$$
Conversely, given a tuple $(b_1,b_2,\ldots,b_m)$ with $1\leq b_1\lt b_2\lt\cdots\lt b_m\leq n+m-1$, we have $b_i\geq n+i$; so this tuple "comes from" the tuple $(b_1,b_2-1,b_3-2,\ldots,b_m-m+1)$, which is a nondecreasing tuple with values between $1$ and $n$; i.e., it is one of our original $(a_1,\ldots,a_n)$s.
That means that counting the nondecreasing tuples $(a_1,\ldots,a_m)$ with $1\leq a_1\leq\cdots\leq a_m\leq n$ is equivalent to counting the number of strictly increasing tuples $(b_1,b_2,\ldots,b_m)$ with $1\leq b_1\lt b_2\lt\cdots\lt b_m\leq n+m-1$. This is, in turn, equivalent to counting the number of ways of selecting $m$ objects from among $\{1,2,\ldots,n+m-1\}$, with no repetitions allowed. This is just $\binom{n+m-1}{m}$. 
