I Why did I lose 60% on my proof of Generalized Vandermonde's Identity?

[12john]

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.

 

[mfb]

Can't read the mind of the person who graded it, but you could have made clearer why (or at least mention that) the sum catches every possible way to select m elements. You only documented that every "element" of the LHS is part of the RHS, i.e. LHS <= RHS.

 

[fresh_42]

I miss a comment on the summation. And the product is only implicitly explained. Where is the double counting? I would have expected an argument $\sum \ldots = ?$ but you have looked at the RHS and reasoned from there instead of the other way around.

 

[vela]

To put what the others said in another way, the way I read your proof, you seem to be saying
$$\binom{n_1}{k_1}\binom{n_2}{k_2} \cdots \binom{n_p}{k_p} = \binom{n_1+\cdots+n_p}{m}$$ as long as $m=k_1+\cdots+k_p$.