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For ex. $\binom{n}{n}$ = $\binom{n-1}{n-1}$ + $\binom{n-1}{n}$ according to the rule $\binom{n}{i}$ = $\binom{n-1}{i-1}$ + $\binom{n-1}{i}$

RM777
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  • That is done by convention. – Bernard Nov 04 '18 at 14:17
  • Ok this is helpful for induction proofs – RM777 Nov 04 '18 at 14:18
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    Yes, it is. You can also go for $\binom{n}{k}$ defined for $n\in\mathbb Z_{\geq0}$ and $k\in\mathbb Z$ where $\binom{n}{k}=0$ if $k\notin{0,\dots,n}$. It is advisable though to mention it if you use that convention. – drhab Nov 04 '18 at 14:26

2 Answers2

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Yes, and it's not an arbitrary "convention". By definition, $$\binom x{n+1}=\frac{x(x-1)(x-2)\cdots(x-n)}{(n+1)!},$$ a polynomial of degree $n+1$ with zeros at $x=0,1,2,\dots,n$.

bof
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Defining $\binom{n}{k}$ as the number of size-$k$ subsets of a size-$n$ set, your statement is indeed correct. You can even keep the usual formula in terms of factorials, viz. $\frac{1}{(-1)!}=\frac{1}{\infty}=0$.

J.G.
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