Defining binomial coefficients over rings in general

In the formula, $\binom{n}{k}$ is an ordinary integer. It need not be viewed as an element of our ring $A$, and in general commutative rings, which may not have a multiplicative unit, it cannot be so viewed.

If $r$ is a ring element, then $\binom{n}{k}r$ is interpreted as the sum in the ring of $\binom{n}{k}$ copies of $r$.

Although the expression ${n \choose k} = \frac{n!}{(n-k)!k!}$ involves division, it actually always evaluates to be an integer (namely, the cardinality of the set of $k$-element subsets of a set with $n$ elements). So in a general ring, ${n\choose k}$ just refers to that integer in that ring (even though the formula $\frac{n!}{(n-k)!k!}$ may not make sense).

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).