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These binomial equations are all true.

$\binom{7}{4} = \binom{4}{4} + 3[ \binom{4}{3} + \binom{4}{2}] + \binom{4}{1}$

$\binom{8}{4} = \binom{5}{4} + 3[ \binom{5}{3} + \binom{5}{2}] + \binom{5}{1}$

$\binom{9}{4} = \binom{6}{4} + 3[ \binom{6}{3} + \binom{6}{2}] + \binom{6}{1}$

$\dots$

Is this the result of some general binomial identity or theory?

My work

I stumbled upon this while working with a recurrence relation in $(m,n)$ with $m,n \in \Bbb Z^+$
that I was trying to put into closed form.

CopyPasteIt
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1 Answers1

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By repeatedly applying Pascal's Theorem:

$$\begin{align*} \binom{n}{4} &=\binom{n-1}{4}+\binom{n-1}{3} \\ &=\left(\binom{n-2}{4}+\binom{n-2}{3}\right)+\left(\binom{n-2}{3}+\binom{n-2}{2}\right) \\ &=\left(\left(\binom{n-3}{4}+\binom{n-3}{3}\right)+\left(\binom{n-3}{3}+\binom{n-3}{2}\right)\right)+\left(\left(\binom{n-3}{3}+\binom{n-3}{2}\right)+\left(\binom{n-3}{2}+\binom{n-3}{1}\right)\right) \\ &= \binom{n-3}{4}+3\binom{n-3}{3}+3\binom{n-3}{2}+\binom{n-3}{1} \end{align*} $$

Sudix
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