Combinatorial proof of this identity: $ \sum_{k=0}^m \left(\!\!\binom{n}k\!\!\right) = \left(\!\!\binom{n+1}m\!\!\right), n\geq0$

Asked Jul 27 '21 at 05:48

Active Jul 17 '23 at 15:47

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I'm trying to give a combinatorial proof of this identity: $$ \sum_{k=0}^m \left(\!\!\binom{n}k\!\!\right) = \left(\!\!\binom{n+1}m\!\!\right), n\geq0 $$

I think the right-hand side could be interpreted as to cast k votes for elements from the set of candidates $[n+1] = {1, 2, 3, . . . , n+1} $ separately, but I'm having difficulty making a consistent argument to show two sides of the equation are equivalent (instead of using definitions to give a mathematical derivation). Thanks for the help!

Note: $\displaystyle\left(\!\!\binom{n}{k}\!\!\right)$ stands for multiset.

edited Jul 17 '23 at 15:47

user1176409

asked Jul 27 '21 at 05:48

IGY

Unless the double parenthesis means something other than binomial coefficients, I don't see how this identity is true. – Alan Abraham Jul 27 '21 at 05:51
@Alan Abraham That's multiset, I'm not sure how to directly type that – IGY Jul 27 '21 at 05:56

Combinatorial proof of this identity: $ \sum_{k=0}^m \left(\!\!\binom{n}k\!\!\right) = \left(\!\!\binom{n+1}m\!\!\right), n\geq0$

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