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I know that $$\binom{n}{k} = \binom{n}{n-k}$$

My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?

An example: Can you define $m$ in terms of $n$ and $k$ such that $$\binom{n}{k}=\binom{n+m}{k}$$

Shaun
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Shrey Joshi
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    No, because, for instance, $$5={5\choose 1} \neq {5+m\choose 1}=5+m$$ for any $m\neq 0$. This becomes obvious once you review the construction of pascal's triangle. – David P Nov 24 '18 at 03:12

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It is standard to define $$ \binom{-n}{k}=(-1)^k\binom{n+k-1}{k} $$ as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)

Micah
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