Let $(c_{n,k})_{n,k=0}^{\infty}$ be defined by
$c(0,0)=1$, $\:\:c(0,k)=0 \:\: \forall \: k > 0$
$$c(n,k) = c(n-1,k) + c(n-1,k-1) \:\:\: \forall \: n \geq 1$$
I can show that the answer is
$$c(n,k)=\frac{n(n-1) \cdots (n-k+1)}{k!} = \binom{n}{k}$$
by recognizing that the binomial coefficients satisfy exactly
the same recurrence but is there another way at arriving at
this result by "algebraic methods"? I am just curios if there is.