# Simplification of binomial coefficient

Is it possible to separate this expression into two expressions each missing one of the variables? $$\binom{m}{n+k} \overset{?}{=}f(m,k) \cdot g(m,n)$$

Edit: The operation can be $$+$$ as well, if that's possible.

• Do you want it with proof? Do you mind division? – NoChance May 20 at 15:34
• Division is fine as well; I just need to separate one of these outside a summation with other multiplicative terms. Proof would be nice :) If not, I can try proving it myself as well. – SS_C4 May 20 at 15:39
• Identity #134 Page 67 here may help. books.google.com.eg/… – NoChance May 20 at 15:53
• The best I can think of is $$\binom{m}{n+k}=\sum_{i=0}^{m-1} \binom{i}{n}\binom{m-1-i}{k-1}.$$So I cannot quite separate $k$ from $n$, but I can write it as a sum of terms which are separated. For a proof, see math.stackexchange.com/questions/1938753/…. – Mike Earnest May 20 at 16:11
• @NoChance, I don't quite see how it helps; could you show me? – SS_C4 May 20 at 16:52

$$\binom{m}{n+k} = f(m,k) \cdot g(m,n)$$
Then $$f(3,2) g(3,2) = 0$$, but $$f(3,2) g(3,1) = f(3,1) g(3,2) = 1$$. QED
If the operation is changed to addition, we have a similar contradiction based on $$m=3$$, $$(n, k) \in \{1,2\} \times \{1,2\}$$.
• How on Earth do you want $f(m,k)$ to only be defined when $n + k \le m$? It is, by design, independent of $n$. – Peter Taylor May 21 at 14:48