Proof that $\sum_{k=1}^{n}{k^{3}\binom{n}{k}^{2}}=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}$

Question

I want to prove that the following equation is correct:

$$
\sum_{k=1}^{n}{k^{3}\binom{n}{k}^{2}}=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}
$$

I used combinatorial proof. There are two companies $A$ and $B$ and each have $n$ employees. Company $A$ and $B$ swap some employees but the number of employees in each companies remain the same. Afterwards company $A$ elect a technical manager and company $B$ elect a technical manager and marketing manager, can be the same person. However, manager in $A$ used to be in $B$ and manager(s) in $B$ used to be in $A$.

Method 1

Select $k$ employees from each company to be swapped, $k\in\{1,2,…,n\}$. From $k$ ex $B$ select one technical manager and from $k$ ex $A$ select technical and marketing manager. The number of possibilities is given below

$$
\sum_{k=1}^{n}{\binom{n}{k}^{2}\cdot k\cdot k^{2}}
$$

Method 2

Select two people from $A$ to be marketing and technical manager respectively then move them to $B$. Select one people from $B$ to be technical manager and move them to $A$. Then complete the companies to make a total of $n$ employees per employees. The number of possibilities is given below

$$
2\binom{n}{2}\cdot\binom{n}{1}\cdot\binom{2n-3}{n-2}
$$

If technical and marketing manager is the same person, select one person from $A$ to be both and move him to $B$. Select one person from $B$ to be technical manager and move him to $A$. Then complete the company member. The number of possibilities is given below

$$
n\cdot n\cdot\binom{2n-2}{n-1}
$$

Since I counted the same possibilities using different methods, the two expressions must be equal.

I want to know if my solution is correct and if anyone has alternative e.g. generating function or algebra solution

Answer 1

Your proof is fine. Here it is a direct approach. Note that $$k^3=k(k-1)(k-2)+3k(k-1)+k.$$ Hence, by Vandermonde's identity, we have $$\begin{align} \sum_{k=1}^{n}k^{3}\binom{n}{k}^{2}&=\sum_{k=1}^{n}k(k-1)(k-2)\binom{n}{k}\binom{n}{n-k}+3\sum_{k=1}^{n}k(k-1)\binom{n}{k}\binom{n}{n-k}\\ &\qquad\qquad +\sum_{k=1}^{n}k\binom{n}{k}\binom{n}{n-k}\\ &=n(n-1)(n-2)\sum_{k=3}^{n}\binom{n-3}{k-3}\binom{n}{n-k}+3n(n-1)\sum_{k=2}^{n}\binom{n-2}{k-2}\binom{n}{n-k}\\ &\qquad\qquad +n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k}\\ &=n(n-1)(n-2)\binom{2n-3}{n-3}+3n(n-1)\binom{2n-2}{n-2}+n\binom{2n-1}{n-1}\\ &=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}. \end{align}$$