scherz0 Posted January 29, 2022 Share Posted January 29, 2022 (edited) My tests are submitted and marked anonymously. I got a 2/5 on the following, but the grader wrote no feedback besides that more detail was required. What details could I have added? How could I perfect my proof? Prove Generalized Vandermonde's Identity, solely using a story proof or double counting. DON'T prove using algebra or induction — if you do, you earn zero marks. $\sum\limits_{k_1+\cdots +k_p = m} {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p} = { n_1+\dots +n_p \choose m }.$ Beneath is my proof graded 2/5. I start by clarifying that the summation ranges over all lists of NONnegative integers $(k_1,k_2,\dots,k_p)$ for which $k_1 + \dots + k_p = m$. These $k_i$ integers are NONnegative, because this summation's addend or argument contains $\binom{n_i}{k_i}$. On the LHS, you choose $k_1$ elements out of a first set of $n_1$ elements; then $k_2$ out of another set of $n_2$ elements, and so on, through $p$ such sets — until you've chosen a total of $m$ elements from the $p$ sets. Thus, on the LHS, you are choosing $m$ elements out of $n_1+\dots +n_p$, which is exactly the RHS. Q.E.D. Edited January 29, 2022 by scherz0 Link to comment Share on other sites More sharing options...
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