$$\sum_{i=k}^{n-l} {i \choose k}{n-i \choose l} = {n+1 \choose k+l+1}$$ I understand how to prove this using that equality: $$\sum_{i=k}^{n-l} {i \choose k}{n-i \choose l} = {n \choose k+l} + \sum_{i=k}^{n-l-1} {i \choose k}{n-i-1 \choose l }$$ But I came to it with the help of my intuition, if there is a strict solution, I just want to know about it.
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Off point: Alternative set theory approach is to consider choosing $~(k + l + 1)~$ seats from a row of $~(n+1)~$ seats. From the left, chosen seat $~(k+1)~$ will have $~k~$ chosen seats to its left and $~l~$ chosen seats to its right. Then, the possible positions of chosen seat $~(k+1)~$ is seat number $~(k+1)~$ through seat number $~(n+1-l),~$ inclusive. For each choice $~(i+1)~$ of chosen seat $~(k+1),~$ the number of ways of having $~k~$ chosen seats to its left and $~l~$ chosen seats to its right is $~\displaystyle \binom{i}{k} \times \binom{[n+1] - [i+1]}{l}.$ – user2661923 Dec 23 '23 at 00:00
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thanks your tips really helped me – Arthur_Kitsuragi Dec 23 '23 at 00:04