This question is in the context of shuffling a standard deck of cards.
As we know, in practice a person cutting a deck will cut somewhere near the center, i.e. the two piles of cards being shuffled together will have somewhere in the neighborhood of 26 cards each.
Assuming a specific cut result, for instance 23 cards in pile A and 29 cards in pile B, how many ways can these be riffled together?
Presumably, in a very rare cut of 3 cards in pile A and 49 cards in pile B, there are far FEWER ways these two piles could be riffled than in a more equal cut. But how do I begin coming up for an equation / real numbers?
The simplest case is easy, when the cut result is 0 cards in pile A and 52 cards in pile B (no cut) and also pretty easy when 1 card in pile A and 51 cards in pile B, but I've tried manually working out 2 and 3 cards in pile A and quickly become stuck.
The trickiest bit for me is understanding how to account for the fact that the piles must maintain their ordering in the resultant pile. i.e. if there's a 5 of hearts at the top of pile A and an 8 of clubs at the bottom, the permutations need to take into account the fact that permutations with the 8 of clubs on top of the 5 of hearts are invalid.
Actually the fact that each pile must maintain its order makes this problem easier in a way.
Let's take the example of a 23 / 29 split. To shuffle them together, the only freedom we have is which 23 of the 52 positions in the deck the pile-A cards end up in; we can't decide any more than that, because those 23 cards have to end up in order, and the other 29 cards have to end up in order as well.
In other words, once we have chosen 23 of the 52 positions, that completely determines the shuffled order. And any choice of 23 positions out of 52 does determine a shuffled order (imagine creating that "shuffle" one card at a time according to our chosen positions). So the total number of shuffles is the binomial coefficient $\binom{52}{23}$. (Alternately, you could think of choosing the 29 positions for the pile-B cards—but fortunately it doesn't matter, since $\binom{52}{29} = \binom{52}{23}$!)