Simplification of binomial coefficient
By Sarah Scott •
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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$\begingroup$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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