TITLE: Defining binomial coefficients over rings in general
QUESTION [9 upvotes]: I'm looking over my notes, and we proved that the binomial theorem $$(a + b)^n = \sum_{k = 0}^{n} {n \choose k}a^k b^{n-k}$$ holds on all commutative rings. 
However, I am confused on how we are defining  ${n \choose k}$, since the definition $ {n \choose k} = \frac{n!}{(n-k)!k!} $ is no longer well defined over rings in general, since $k!$ may equal $0$. For example, $p! = 0$ in $F_p$, the field of p elements. Any idea how to interpret the binomial coefficients in these cases?

REPLY [3 votes]: If the division bothers you, it is entirely possible to use a definition for the binomial coefficients that does not involve dividing anything at all. In particular, you can use the recurrence relation $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$ with the starting conditions $$\binom{n}{0}=\binom{n}{n}=1$$ as a definition (which of course is just the mathematical way of presenting the construction of Pascal's triangle).