Binomials in associative algebras Given any associative algebra $A$ over a field of characteristic zero, $x\in A$ and $k\in \mathbb Z_+$, set $\binom{x}{k} = \frac{x(x-1)\cdots(x-k+1)}{k!}$. It is not hard to see that is is still in $A$.
How to express $\binom{x+y}{k}$ in terms of a linear combination of products of $\binom{x}{m}$ and $\binom{y}{n}$ for appropriate $m,n\in \mathbb Z_+$ ? 
Thanks,
 A: The argument I have in mind seems to be engendering some confusion, so I'll spell it out in full. The crucial lemma is the following.
Lemma: Suppose $P(x_1, ... x_n)$ is a polynomial. If $P(k_1, ... k_n) = 0$ for all non-negative integers $k_1, ... k_n$, then $P = 0$ identically.
Proof. We proceed by induction on $n$. The case $n = 1$ is obvious. In the general case, regard $P(x_1, ... x_n)$ as a polynomial in $x_n$ with coefficients in $F[x_1, ... x_{n-1}]$, so write
$$P(x_1, ... x_n) = \sum_i P_i(x_1, ... x_{n-1}) x_n^i.$$
By fixing $x_1, ... x_{n-1}$ and varying $x_n$ we conclude that each of the polynomials $P_i(x_1, ... x_{n-1})$ satisfies $P_i(k_1, ... k_{n-1}) = 0$ for all non-negative integers $k_1, ... k_{n-1}$, and the result follows by induction.
Consider now the polynomial
$$P(x, y) = {x + y \choose m} - \sum_{a=0}^m {x \choose a} {y \choose m-a}.$$
The combinatorial argument in Dimitrije Kostic's answer shows that $P(k_1, k_2) = 0$ for all non-negative integers $k_1, k_2$, and then the lemma above shows that $P = 0$ identically. In other words, $P(x, y) = 0$ in the universal commutative $\mathbb{Q}$-algebra generated by two generators $\mathbb{Q}[x, y]$, and so $P = 0$ for elements $x, y$ in any commutative $\mathbb{Q}$-algebra. 
A: There's a simple combinatorial formula.  For all positive integers $x,y,m$ you have
$$ \binom{x+y}{m} = \sum_{l=0}^m \binom{x}{l} \binom{y}{m-l}
$$
It's easy to prove.  If you're choosing $m$ elements from a set of $x+y$ elements, you can do this by taking some number $l$ of them from among the first $x$ elements and then the remaining $m-l$ from the last $y$ elements.  Could this be what you're looking for?
