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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.

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  • $\begingroup$ Do you want it with proof? Do you mind division? $\endgroup$ – NoChance May 20 at 15:34
  • $\begingroup$ 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. $\endgroup$ – SS_C4 May 20 at 15:39
  • $\begingroup$ Identity #134 Page 67 here may help. books.google.com.eg/… $\endgroup$ – NoChance May 20 at 15:53
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    $\begingroup$ 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/…. $\endgroup$ – Mike Earnest May 20 at 16:11
  • $\begingroup$ @NoChance, I don't quite see how it helps; could you show me? $\endgroup$ – SS_C4 May 20 at 16:52
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No, it can't be done. Suppose

$$\binom{m}{n+k} = f(m,k) \cdot g(m,n)$$

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 $m=3$, $(n, k) \in \{1,2\} \times \{1,2\}$.

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  • $\begingroup$ Suppose the functions are only supposed to be defined when n+k <= m. I'm not very comfortable with having \binom(3,4) = 0. $\endgroup$ – SS_C4 May 21 at 14:38
  • $\begingroup$ How on Earth do you want $f(m,k)$ to only be defined when $n + k \le m$? It is, by design, independent of $n$. $\endgroup$ – Peter Taylor May 21 at 14:48

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