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This is an inverted binomial table:

$$ \begin{array}{ccccccccccc} &&&&&&1&&&&&&\\ &&&&&1&&1&&&&&\\ &&&&2&&1&&2&&&&\\ &&&6&&2&&2&&6&&\\ &&24&&6&&4&&6&&24&&\\ &120&&24&&12&&12&&24&&120\\ 720&&120&&48&&36&&48&&120&&720 \end{array} $$

Each element $(n,k)$ is $\dfrac{n!}{\dbinom{n}{k}}=(n-k)!k!$.

Can we prove $(n,k)=\dfrac{(n-1,k)(n-1,k-1)}{(n-2,k-1)}$?

JMP
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1 Answers1

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Writing in terms of factorials, $$RHS=\frac{(n-1-k)!\,k!\,(n-k)!\,(k-1)!}{(n-1-k)!\,(k-1)!}=(n-k)!\,k!=LHS\ .$$

Alternatively, from the well known relation $$k\binom nk=n\binom{n-1}{k-1}$$ we get $$\frac{n!}{k\binom nk}=\frac{n!}{n\binom{n-1}{k-1}}\ ,$$ that is, $$(n,k)=k(n-1,k-1)\ .$$ Hence $$\frac{(n,k)}{(n-1,k-1)}=\frac{(n-1,k)}{(n-2,k-1)}$$ since each side equals $k$, and your identity follows immediately.
David
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