1

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.

SS_C4
  • 1,037
  • Do you want it with proof? Do you mind division? – NoChance May 20 '19 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 '19 at 15:39
  • Identity #134 Page 67 here may help. https://books.google.com.eg/books?id=ovUwGLm2IJEC&pg=PA63&dq=binomial+identities&hl=en&sa=X&ved=0ahUKEwizydLetariAhVmx4UKHZYSD54Q6AEILTAB#v=onepage&q=binomial%20identities&f=false – NoChance May 20 '19 at 15:53
  • 1
    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 https://math.stackexchange.com/questions/1938753/sum-k-mn-binommkr-binomn-ks-binommn1rs1-using-coun/3193799. – Mike Earnest May 20 '19 at 16:11
  • @NoChance, I don't quite see how it helps; could you show me? – SS_C4 May 20 '19 at 16:52
  • @MikeEarnest, That's correct, but it doesn't help me answer my question :( – SS_C4 May 20 '19 at 16:53
  • Maybe you could manipulate the identity to fit your need. I don't know how to, it is one of the few identities with 3 variables and without a sum. – NoChance May 20 '19 at 22:29

1 Answers1

0

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

Peter Taylor
  • 13,701
  • Suppose the functions are only supposed to be defined when n+k <= m. I'm not very comfortable with having \binom(3,4) = 0. – SS_C4 May 21 '19 at 14:38
  • 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 '19 at 14:48