I wish to compute the sum \begin{align} \sum_{i_1,i_2,\dots,i_{2k}=1}^n1 \end{align} with restriction that at least for one fixed index $i_s$ we have $i_s\neq i_r$ if $r\neq s$.
This arises from trying to count the number of terms that are zero in the following expectation for centred i.i.d random variables $X_i$ \begin{align*} \mathbb{E}(\sum_{i=1}^nX_i)^{2k}=\sum_{i_1,i_2,\dots,i_{2k}=1}^n\mathbb{E}(\prod_{s=1}^{2k}X_{i_s}) \end{align*} or equivalently \begin{align*} \mathbb{E}(\sum_{i=1}^nX_i)^{2k}=\sum_{\substack{k_1+k_2+\dots+k_n=2k\\ k_i\geq0 \forall i=1,2,\dots,n}}\binom{2k}{k_1,k_2,\dots k_n}X_i^{k_1}X_2^{k_2}\dots X_n^{k_n}. \end{align*} Thus equivalently I could also compute \begin{align*} \sum_{\substack{1+k_2+\dots+k_n=2k\\ k_i\geq0 \forall i=2,\dots,n}}&\binom{2k}{1,k_2,\dots k_n}+\sum_{\substack{k_1+1+\dots+k_n=2k\\ k_i\geq0 \forall i=1,3,\dots,n}}\binom{2k}{k_1,1,\dots k_n}+\dots \\ &+\sum_{\substack{k_1+k_2+\dots+1=2k\\ k_i\geq0 \forall i=1,2,\dots,n-1}}\binom{2k}{k_1,k_2,\dots 1}, \end{align*} but apart from obvious symmetry in the terms I don't know how to start evaluating this either.
Edit. As I have commented below what I'm actually after is the sum
\begin{align*} \sum_{\substack{k_1+k_2+\dots+k_n=2k} \\ \,\,\,\, \text{even partitions}}\binom{2k}{k_1,k_2,\dots,k_n}. \end{align*}
The lowest sum above in my original question is actually computable, but I don't think it is correct for the original question.
we know by taking $x_i=1$ for all $i$ in the multinomial theorem that
\begin{align*}
\sum_{\substack{k_1+k_2+\dots+k_n=m}\\ \,\,\,\,k_1,k_2,\dots,k_n\geq 0}\binom{m}{k_1,k_2,\dots,k_n}=n^m.
\end{align*}
Also noting that
\begin{align*}
\binom{2k}{k_1,\dots,k_i=1,\dots,k_n}&=\binom{2k}{k_1=1,k_2,\dots,k_n}=\dfrac{(2k)!}{1!k_2!\dots k_n!}=2k\dfrac{(2k-1)!}{k_2!\dots k_n!} \\
&=2k\binom{2k-1}{k_1,k_2,\dots,k_{n-1}},
\end{align*}
where the last step is just a relabelling the indices. Then
\begin{align*}
\sum_{\substack{1+k_2+\dots+k_n=2k\\
k_i\geq0 \forall i=2,\dots,n}}&\binom{2k}{1,k_2,\dots k_n}+\sum_{\substack{k_1+1+\dots+k_n=2k\\
k_i\geq0 \forall i=1,3,\dots,n}}\binom{2k}{k_1,1,\dots k_n}+\dots \\
&+\sum_{\substack{k_1+k_2+\dots+1=2k\\
k_i\geq0 \forall i=1,2,\dots,n-1}}\binom{2k}{k_1,k_2,\dots 1}=2kn(n-1)^{2k-1},
\end{align*}
which seems too large to be the correct answer to the original question.
Perhaps some inclusion-exclusion argument is needed here and also in the question of the even partitions.
For a fixed $m=2s-1<2k$ and $j\in\{1,2,\dots,n\}$ we can make similar computation
\begin{align*} \sum_{\substack{k_1+\dots+k_j=m+\dots+k_n=2k}\\ \,\,\,\,\,\,k_1,k_2,\dots,k_n\geq 0}&\binom{2k}{k_1,\dots,k_j=m,\dots,k_n}\\ &=\dfrac{2k(2k-1)\dots(2k-m)}{m!}\sum_{\substack{k_1+k_+\dots+k_n=2k-m}\\ \,\,\,\,\,\,k_1,k_2,\dots,k_n\geq 0}\binom{2k-m}{k_1,k_2,\dots,k_{n-1}}\\ &=\dfrac{2k(2k-1)\dots(2k-m)}{m!}(n-1)^{2k-m} \\ &=\dfrac{2k(2k-1)\dots(2k-2s+1)}{(2s-1)!}(n-1)^{2k-2s+1} \end{align*}
Now for each $s\leq k$ we get $n$ such terms, but we should sum over all $s$ with appropriate "weighting" to avoid overcounting or something like this.
Where does my reasoning go wrong here?
