I am having trouble creating a proof for the below equation to show that they are in fact equal to one another. n and k are positive integers.

$$ \left(\!\!{n\choose k}\!\!\right) = \sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right)$$

Any help would be greatly appreciated!

Overall I have tried multiple ways to relate it to something similar like the below to help get closer to something less complex, $$\sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right) = {n+k -1\choose k}$$

however I can't seem to get anywhere that makes any sense in regards to proving the two first equations are equal to one another.


One approach is to argue combinatorially. Suppose that you want to choose a $k$-element multiset from the set $[n]=\{1,\ldots,n\}$. Let $\mathscr{M}$ be the family of all such multisets, and for $\ell=0,\ldots,k$ let $\mathscr{M}_\ell$ be the set of multisets in $\mathscr{M}$ that have $\ell$ copies of $n$. If $M\in\mathscr{M}_\ell$, then what remains of $M$ when you throw out the copies of $n$ is a multiset of cardinality $k-\ell$ chosen from the set $[n-1]$, so


and of course


Can you finish it from here?

Another possibility is to reduce it to binomial coefficients and try to show that


This can be rewritten as


which is sometimes known as the hockey stick identity and has several proofs here.


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