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Pascal's identity states that $\binom{n}{k}$ = $\binom{n-1}{k-1}$ + $\binom{n}{k+1}$

However, if we let k = n then according to the identity we have that $\binom{n-1}{n-1}$ + $\binom{n}{n+1}$ which is then undefined if I am correct. So must n>k for the identity to be defined?

RandomUser
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    Yes or you assume that ${a\choose b}=0$ when $b>a$ or $b<0$ as it is typically done by convention. When $b>a$ or $b<0$ then the number of $b$-subsets of an $a$-set is indeed $0$ so it all works. Then it works for all $n,k$. – Michal Adamaszek Jun 21 '22 at 10:29
  • @MichalAdamaszek Thanks. Maybe you can post an answer so I can bestow you the honor of solving it. – RandomUser Jun 21 '22 at 10:32
  • @MichalAdamaszek There is a slight issue (not relevant here) about ${a \choose b}$ with negative $a$ and positive integer $b$, particularly when $0 \le b \le -a$ – Henry Jun 21 '22 at 11:08

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