I am trying to come up with a combinatorial proof of the following fact:
$$ \sum_{k_1+...+k_m=n}{n \choose k_1...k_m}(-1)^{k_2+k_4+...+k_{2l}}= \begin{cases} 0 & \text{if $m=2l$} \\ 1 & \text{if $m=2l+1$} \end{cases} $$
This is a special case of the multinomial theorem in which the $x_i$'s alternate between $1$ and $-1$. I'm wondering if it also has a combinatorial interpretation/proof.
In the $m=2$ case, if alternating entries of each row of Pascal's triangle are multiplied by -1, then the sum of each row will be $0$. (For instance, the fifth row becomes $1-4+6-4+1=0$, and the sixth row becomes $1-5+10-10+5-1=0$.) Is there a way to extend this idea to the general case? Thanks in advance to anyone who's able to help.
First, let us prove the $m=2l$ case.
Note that $\binom{n}{k_1\ldots k_{2l}}$ counts the number of ways to label $a_1,\ldots, a_n$ with the colors $1,2,\ldots, 2l$ such that each color $i$ is given to $k_i$ elements. Let us say such a labeling is an extra-even labeling if there are an even number of even labels (i.e. $(-1)^{k_2+\ldots + k_{2l}}=1$). The identity is the same as showing that the number of extra-even labelings is the same as the number of non-extra-even labelings; that is, the labelings with an odd number of even labels (i.e. $(-1)^{k_2+\ldots + k_{2l}}=-1$).
To do this, we define a permutation of all labelings as follows: if $a_1$ is given an even label $2x$, we relabel it to the label $2x-1$. If it is given an odd label $2x-1$, we relabel it to $2x$. This changes the total number of even labels by 1, hence gives a bijection between the sets of extra-even labelings and non-extra-even labelings, hence the two sets are the same size.
The same strategy works for $m=2l+1$. We just change the permutation as follows: we fix the unique labeling where all labels are $2l+1$; otherwise, if $p$ is the first index so that $a_p$ is not labeled by $2l+1$, we do the above label swapping of $2x\leftrightarrow 2x-1$ for $a_p$. This again gives the correspondence between extra-even labelings and non-extra-even labelings, with the exception of the fixed point, which is extra-even. Thus, the number of extra-even labelings in this case is the same as the non-extra-even labelings plus one extra.