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This might very well have already been asked and answered before but I did not find this question in that form. So, I wonder if asking

how many $m$-elements subsets are contained in an $n$-elements set?

is equivalent to asking how many ways can I choose $m$ from $n$, provided that $n\ge m$?

I know this is a trivial question but I want to make sure I understand the basics correctly before moving on to topics of combinatorial reasoning to approach number theory.

So we are looking for all sets $A_x$ such that $|A_x|=m$, that can be chosen from a set $S$ such that $|S|=n$. The way I see it, since order does not matter and since we do not count a repeated element twice, what I get to is a simple combination, $$\frac {n!}{(n-m)!\,m!}\;=\;\binom nm$$

I noticed that this also seems equivalent to a basic committee problem. If I want to form a $m$-persons committee from $n$-persons such that $n\ge m$, what we want is to find a set $C$ such that $|C|=m$ from a set of persons $S$ such that $|S|=n$.

Again, I think the order in which I select the $n$-persons is not important, since choosing Jane before John does not change $|C|$. Also, we are selecting persons, so it follows that repetition is not allowed since to form a committee of $n$ members we must select $n$ different persons. So again, I find that there are $$\frac {n!}{(n-m)!\,m!}\,=\,\binom nm$$ ways to select $C$ from $S$.

Finally, is this also the same question as to ask what is the coefficient of $a^mb^r$ in the expansion of $(a+b)^n$? From the binomial theorem I would say the answer is also $\binom nm$ such that $m=n-r$.

These questions seem all like the same to me, is this correct or is this the wrong way to think about this kind of problems?

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    Yes, this is precisely the meaning of $m \choose n$. – 2017-02-22
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    Yes, that is correct. That is why the binomial is sometimes called "$m$ choose $n$" because it equals the number of ways to choose a size-$n$ subset of a size-$n$ set. – 2017-02-22
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    $n \le m$ it is. – 2017-02-22
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    n choose m subsets – 2017-02-24

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After further reading I feel confortable answering my own question. As pointed out in the comments this is exactly the definition of combinations.

So, first of all, a $r$-permutation of a set $S$ containing $n$ elements is a an ordered selection from $S$, containing $r$ elements. We have $n$ possibilities for our first selection, then $(n-1)$ for the next selection and so on.

Thus, it can be shown using the principle of mathematical induction that the $i^{th}$ selection can be made in exactly $n-(i-1)$ ways such that, from the general combinatorial principle, the total number denoted $_nP_r$ of possible $r$ elements ordered selections that can be made from a set $S$ containing $n$ elements is $$_nP_r\,=\,n(n-1)(n-2)\ldots(n-r+1)\,=\, \frac{n!}{(n-r)!}$$ A $r$-combination from a set $S$ containing $n$ elements is a subset of $S$ containing $r$ elements. The number $_nC_r$ of possible $r$ elements subsets of a set $S$ containing $n$ elements is $$_nC_r\,=\,\binom nr \,=\, \frac {n(n-1)(n-2) \ldots (n-r+1)}{r!} = \frac {_nP_r}{r!}$$ We can make $_rP_r = r!$ different ordered selections of $r$ elements from a set containing $r$ elements. In a combination, the order of the elements do not matter. Hence we get the number of $r$ elements subsets of the set $S$ containing $n$ elements, by dividing the number of possible $r$ elements ordered selections from the set $S$ containing $n$ elements by the number $_rP_r$ of different ways we can order such $r$ elements selection.

Therefore, $$_nC_r\,=\,\binom nr \,=\, \frac{_nP_r}{_rP_r}\,=\, \frac{n!}{(n-1)!\,r!}$$ Finally, we see that since $_nP_r$ contains all possible orderings of $r$ elements selections from the set $S$ containing $n$ elements, it follows that this number is greater than the number of possible different $r$ elements subsets of $S$ such that $_nP_r \,\gt\, _nC_r$.