Let us number our objects $1,2,\ldots,n$. Any selection of $m$ elements from these $n$ possibilities with repetition can be described as an $m$-tuple where the entries are non-decreasing: $(a_1,a_2,\ldots,a_m)$, with $1\leq a_1\leq a_2\leq\cdots\leq a_m\leq n$. This expression is unique.
Now consider the tuple $(b_1,\ldots,b_m)$ obtained from $(a_1,\ldots,a_m)$ by letting
$$(b_1,\ldots,b_m) = (a_1,a_2+1,a_3+2,\ldots,a_m+(m-1)).$$
Notice that $1\leq b_1\lt b_2\lt\cdots\lt b_n\leq n+m-1$; moreover, distinct $a$-tuples correspond to distinct $b$-tuples; and, more importantly, every $m$-tuple $(c_1,\ldots,c_m)$ with $1\leq c_1\lt c_2\lt\cdots\lt c_m\leq n+m-1$ corresponds to an $a$-tuple, namely, $(c_1,c_2-1,\ldots,c_m-m+1)$ (which will satisfy $1\leq c_1\leq c_2-1\leq\cdots\leq c_m-m+1\leq n$).
Thus, counting $a$-tuples (that is, combinations with repetitions from $\{1,\ldots,n\}$) is equivalent to counting $b$-tuples; the advantage is that to count $b$-tuples we just need to count the number of possible $m$-tuples chosen from $\{1,2,\ldots,n+m-1\}$ without replacement. This is the basic formula $\binom{n+m-1}{m}$. Thus, the number of possible combinations with repetitions of $m$ elements chosen from $n$ possibilities is
$$\binom{n+m-1}{m}.$$
Consider the set $\{1,\ldots,n\}$. Add in $m-1$ new symbols, $r_1,\ldots,r_{m-1}$. Think of $r_i$ as "repeat the $i$th symbol."
Now choose without repetition an $m$-tuple from $\{1,\ldots,n,r_1,\ldots,r_{m-1}\}$. Write it out in the order that has every $r$ larger than every number, the numbers ordered in their usual way, and the $r$s ordered by their indices. For example, you might get $2,3,r_1,r_3,r_4$. This will correspond to the $m$-tuple-with-repetitions obtains by replacing $r_i$ with whatever is in the $i$th position, hence here we get
$$2, 3, 2, 2, 2$$
You'll want to convince yourself here as well that every $m$-tuple-with-repetitions from $\{1,2,\ldots,n\}$ corresponds to a single $m$-tuple-without-repetitions from $\{1,2,\ldots,n,r_1,\ldots,r_{m-1}\}$ and vice-versa, so that the number of combinations-with-repetitions from $\{1,2,\ldots,n\}$ is equal to the number of combinations-without-repetitions from $\{1,2,\ldots,n,r_1,\ldots,r_{m-1}\}$. There are $n+m-1$ objects in the latter set, so we again get
$$\binom{n+m-1}{m}.$$
Best Answer
Lets say you have $n$ elements and some particular $m$ objects are fixed (they must be chosen).
There are a total of $c$ elements that need to be chosen with $m$ elements that must be there.
Now this means that there are $c-m$ elements that must be selected from $n-m$ elements.
$$\binom{n-m}{c-m}$$