I hope I successfully used the title to attract all people who might know the answer to my question...

I know basic things like P(n,n), P(n,r), C(n,r)...And I even know the permutation of n choose n where p are alike and q are alike (n!/p!q!). The problem comes up with things like P(n,r) where p are alike and q are alike (as well as the combination equivalent). This comes up a lot and I have no idea how to set that up.

I find it important in card games where suit is irrelevant and only numbers matter. You have 13 sets of 4 alike cards. So as long as the r you're choosing is less than four in this case, you can just have P(n,r) as 13^r and I reasoned without myself that choosing two from the deck in which order does not matter as tri(13).*

*Not sure how it's commonly notated but I've used "tri(n)" here to mean the triangular number of n. The triangular number is the sum of all numbers between 1 and n, or the additive equivalent of factorial. It is most easily found by (n[n+1])/2.

Edit: I found online that C(n,r) with repititions can be represented by (n+r-1)!/r!(n-1)!, but I think this allows repeating a selection for as many times up to R. But we are limited in my card scenario to repeating a selection only as many of four times. So it's not much of a solution when r>4.