An interesting question on the bird site:

How do we know, arithmetically, that nCk is an integer? Why should both k! and (n-k)! simultaneously divide into n! ?

@ColinTheMathmo
Cancel the (n-k)!, and we're left with "why does k! divide the falling factorial $$n^{\underline{k}}$$?" (notation aside). And that can be answered, I think, with an understanding of divisibility by prime powers in ranges of consecutive integers.

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@bmreiniger You can show that each factor in the denominator divides the numerator, but showing that they all do so without fatally interfering with each other is a little trickier.

@ColinTheMathmo @bmreiniger that was my personal stumbling block with the factorial definition. You can divide $$n!$$ by $$k!$$, and you can divide $$n!$$ by $$(n-k)!$$, but being able to divide $$n!$$ by both... that was the surprise.

@ColinTheMathmo (cc @tpfto)
Trickier, sure, but I think my vague approach above works. Lemme try to describe some more (maybe a proofinatoot later); it's similar to the classic exercise that shows that the power of prime $$p$$ in $$k!$$ is $$\sum_{i\geq1} \left\lfloor\frac k{p^i}\right\rfloor$$ \) (which I now realize has a name, "Legendre's formula"). The idea extends to say that the power of prime $$p$$ in the falling factorial $$n^{\underline{k}}$$ is _at least_ that quantity.

@ColinTheMathmo
To be clear, I agree with the other thread here, that the other definitions are more natural. I'm approaching this from "how can we show this directly?".

@bmreiniger I agree that this approach can be pushed through, but it's not immediately obvious, and does require care. The question then is: "How clear *can* this be made?"

The discussion on Twitter has been interesting:

solipsys.co.uk/Chitter/Binomia

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