It seems there should be a way to "directly" determine the number of distinct subsets of a given size $r$ possible using $n$ distinct elements, i.e. number of combinations.
The usual way to think about it is to first determine the number of distinct ordered lists of a size $r$ using $n$ distinct elements (i.e. number of permutations, $n\mathrm{P}r = \frac{n!}{(n-r)!}$), and because that includes different orderings of the same elements, then divide by $r!$ (i.e. the number of permutations of those $r$ elements), giving $n\mathrm{C}r = \frac{n!}{r!(n-r)!}$.
For example, given $6$ elements $a,b,c,d,e,f$, the number of combinations of $3$ of them is the number of permutations of $3$ of them ($6\times 5\times 4 = 120$, or $6\mathrm{P}4 = \frac{6!}{(6-3)!}$). But this includes $abc, acb, bac, bca, cab, cba$ as distinct elements (because distincting orderings of the same three elements). So we divide by the number of ways of arranging $3$ elements (i.e. their permutations, which is $3! = 6$; or using the formula, $ 3\mathrm{P}3 = \frac{3!}{(3-3)!} $ ), which gives $120/6 = 20$.
Is there a way to build up the number of distinct subset directly, instead of working out permutations and then cancelling out the different orderings later?
Thoughts
I think, if there is a "direct" way to think about it, it will be much more complex than this one --- or the above approach wouldn't be taught.
By "direct", I mean something like building a subset (a combination) step by step. But when I do that naively, it includes subsets with their elements in a different order (as in the above): a permutation, not a combination.
A decision tree would show which paths result in the same elements included at different levels. But how to eliminate those from start (instead of cancelling them out later)?
Perhaps it's related to the size of a "powerset", the set of all subsets?
BTW: I realize that any "direct" calculation would be equivalent to the standard one above; I really mean a way to thinking about it/conceptualize it as direct, rather then doing permutations first, then accounting for ordering.
EDIT another way to think of it is as how many $n$ digit binary numbers (including leading $0$'s) have $r$ $1$'s. The $n$ digit binary number is a bitfield (each position represents one of $n$ elements, value of $1$ means it is in the subset, $0$ means it is not); the number of $1$'s ($r$) is how many are in the subset.
Thanks for any insight!
Absolutely! You can build a decision tree out of $_n C _r$ via the identity $$_n C _r = _{n-1}C_r + _{n-1}C_{r-1}.$$ We set out to select a subset $S \subseteq \{1, 2, \ldots, n\}$ of size $r$. Here's the recursive decision tree: