Combinatorics - An unproved slight generalization of a familiar combinatorial identity One of the most basic and famous combinatorial identites is that
$$\sum_{i=0}^n \binom{n}{i} = 2^n \; \forall n \in \mathbb{Z^+} \tag 1$$
There are several ways to make generalizations of $(1)$, one is that:
Rewrite $(1)$: $$\sum_{a_1,a_2 \in \mathbb{N}; \; a_1+a_2=n} \frac{n!}{a_1! a_2!} = 2^n \; \forall n \in \mathbb{Z^+} \tag 2$$
Generalize $(2)$: $$\sum_{a_1,...,a_k \in \mathbb{N}; \; a_1+...+a_k=n} \frac{n!}{a_1!...a_k!} = k^n \; \forall n,k \in \mathbb{Z^+} \tag 3$$
Using Double counting, it is easy to prove that $(3)$ is true. So we have a generalization of $(1)$.
The question is whether we can generalize the identity below using the same idea
$$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n} \; \forall n \in \mathbb{Z^+} \tag 4$$
which means to find $f$ in
$$\sum_{a_1,...,a_k \in \mathbb{N}; \; a_1+...+a_k=n} \left ( \frac{n!}{a_1!...a_k!} \right )^2 = f(n,k) \; \forall n,k \in \mathbb{Z^+} \tag 5$$
That is the problem I try to solve few days now but not achieve anything. Anyone has any ideas, please share. Thank you.
P.S: It is not an assignment, and sorry for my bad English.
Supplement 1: I think I need to make it clear: The problem I suggested is about to find $f$ which satisfies $(5)$. I also show the way I find the problem, and the only purpose of which is that it may provide ideas for solving.
Supplement 2: I think I have proved the identity of $f(n,3)$ in the comment below
$$f(n,3) = \sum_{i=0}^n \binom{n}{i}^2 \binom{2i}{i} \tag 6$$
by using Double Counting:
We double-count the number of ways to choose a sex-equal subgroup, half of which are on the special duty, from the group which includes $n$ men and $n$ women (the empty subgroup is counted).
The first way of counting: The number of satisfying subgroups which contain $2i$ people is $\binom{n}{i}^2 \binom{2i}{i}$. So we have the number of satisfying subgroups is $RHS(6)$.
The second way of counting: The number of satisfying subgroups which contain $2(a_2+a_3)$ people, $a_2$ women on the duty and $a_3$ men on the duty is
$$\left ( \frac{n!}{a_1!a_2!a_3!} \right )^2$$.
So the number of satisfying subgroups is $LHS(6)$.
Hence, $(6)$ is proved.

Induction using Vandermonde's identity
$$
\sum_{i_1+i_2=m}\binom{n_1}{i_1}\binom{n_2}{i_2}=\binom{n_1+n_2}{m}\tag{1}
$$
yields
$$
\sum_{i_1+i_2+\dots+i_k=m}\binom{n_1}{i_1}\binom{n_2}{i_2}\dots\binom{n_k}{i_k}=\binom{n_1+n_2+\dots+n_k}{m}\tag{2}
$$
