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simple binomial theorem proof

Why is ${6\choose 0} + {7\choose 1} + \ldots + {n+6 \choose n} = {n+7 \choose n}\;?$

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    @Pam did you also answer this question: http://math.stackexchange.com/a/262660/19341 with a different account?2012-12-20

2 Answers 2

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HINTS: The algebraic argument is a proof by induction: verify the equality for $n=0$, and show that if ${6\choose 0} + {7\choose 1} + \ldots + {n+6 \choose n} = {n+7 \choose n}\;,$ then ${6\choose 0} + {7\choose 1} + \ldots + {(n+1)+6 \choose {n+1}} = {(n+1)+7 \choose {n+1}}\;.$ This is a pretty straightforward application of the Pascal’s triangle identity.

For the combinatorial argument, observe that $\binom{n+7}n=\binom{n+7}7$ is the number of ways to choose a $7$-element subset of $\{1,2,\dots,n+7\}$, and $\binom{k+6}k=\binom{k+6}6$ is the number of ways to choose a $7$-element subset of $\{1,\dots,n+7\}$ whose largest element is the number $k+1$. That is, to choose a $7$-element subset whose largest element is $10$, you first choose $10$, and then you have to choose $6$ of the numbers $\{1,\dots,9\}$.

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The standard recursive formula for binomial coefficients is

$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

There is a technique for recursion called "unrolling," where you repeatedly substitute a recurrence into its equation for itself. Here if we reverse the terms and then substitute repeatedly we get:

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-2}{k-1} + \binom{n-2}{k-2}$

$\vdots$

$\binom{n}{k} = \binom{n-1}{k} + \binom{n-2}{k-1} + \binom{n-3}{k-2} + \cdots + \binom{n-k}{0}$

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    @Joe: Some people do use the terms that way, but you can’t count on it: many of us use them completely synonymously.2012-12-14