so this is my first time asking a question. I'm currently in my third year of university studying a lot of probability and statistics and even to this day combinatorics has not been my strong suit.
I was wondering if there was an overarching way to interpret a combinatoric or at least a mind set when dealing with these questions. For example $$ C(a,b) = a!/(b!*(a-b)!) $$ So at it's most basic how I learnt this was. C(a,b) is the number of combinations of b items you can get from a items. However when doing probability subjects I've noticed many different ideas are used in the same way. For example. If we have a possible slots for items to go in. C(a,b) is the number of combinations that b items (where a>b) can go into the available a slots.
Or If we have a row of a items, and b of them are identical, and the remaining b - a of them are also identical but not the same as the first b items. C(a,b) is the number of possible permutations of all those items if they were placed in a row.
Honestly wrapping my mind around all these different interpretations gives me a headache. Is there a better overall interpretation/definition of what C(a,b) is and what it represents or is it nothing more than simplified factorical algebra that has too broad an application to generally understand base on application?
By definition, $C(n,k)={n\choose k}$ is the number of $k$-element subsets of the set $[n]:=\{1,2,3,\ldots, n\}$. In order to find a formula for this number one has to do a certain sequence of mental movements, and finally one arrives at $$C(n,k)={n\choose k}={n!\over k!(n-k)!}\qquad(0\leq k\leq n)\ .\tag{1}$$ I'm sure you have seen a proof of this formula, but you would believe it anyway.
But this is only one part of the story. It seems your headaches have to do with something else: Combinatorial problems come in many different guises. You have given several combinatorial stories that all lead to the $C(n,k)$. This as nothing to do with factorials etc., but with structural insight into the problem at hand. E.g., if the given set is a set of $n$ different colors instead of $[n]$ this should not make a difference for the number of color combinations. When you have some exercise you should immediately see that placing $a$ white balls and $b$ black balls in a row amounts to selecting $a$ white places in an empty row of length $a+b$; hence there are ${a+b\choose a}$ ways to do it.