Question in the title. I recently saw a short video on Youtube in which the problem was to evaluate $10!/6!$. The answer of course was $7!$, but that got me wondering: just how many triples $(n,k,c)\in \mathbb{N}^3$ are there such that $n\ne k$ and $k\ne 0$ and $k\ne 1$ and $n!/k!=c!$? How might I go about finding them?
After a few minutes' thought, I realised just how grossly ill-equipped I currently am to answer these questions. The number $n!/k!$ must obviously contain exactly the right number of primes to "replace" those present in the product $k!$, as well as the exact number required to "replace" the composite numbers in that product.
Hence, it seems obvious that the distribution of primes is a critical factor in answering these questions, but that determination is about as far as my current maths ability will take me. You see, I have no formal training in mathematics, so I'm not even sure where to direct my curiosity. Given that the distribution of primes must be considered, I'm fairly sure I'll require some complex number theory, yeah? But what else? I'm drawing a blank, thus this question.
I'm not looking for anybody to solve this for me (though that would be appreciated, provided you provide a full explanation of your result), just for directions on what areas to pursue. Specific resources would be incredibly appreciated. Whether or not this is even something that current mathematics can address is an important question too, I suppose.
Anyway, that's about it. I'm aware that MSE etiquette demands a bit more work than I've done (basically none), but please believe me when I say that I really would have if I could, lol. I'm hoping that you all can forgive that. Cheers for your time :)