It is relatively easy to show that
$$
\sum_{a_1 + \cdots + a_k=\ell} \binom{\ell}{a_1,\ldots,a_k} = k^\ell
$$
where $\binom{\ell}{a_1, \ldots, a_k} = \frac{\ell!}{a_1!\cdots a_k!}$. What can be said if we want to compute the restricted sum
$$
s(\ell,k) = \sum_{a_1 + \cdots + a_k=\ell} \binom{\ell}{a_1,\ldots,a_k}
$$
where we now restrict the summation to those $a_k$ which are odd? At the least, of course, we need that $\ell \geq k$ and that $\ell \equiv k \pmod 2$. Is this sum known in the literature?
The simplest case of $s(2k,2) = 2^{2k-1}$ can be easily verified, but I believe that this is an anomoly based on the fact that these are (secretly) binomial coefficients.
This arises in computing the coefficients of the power series of $\big(\sin(x)\big)^k$.
Best Answer
$\binom{\ell}{a_1,\dots,a_k}$ is the coefficient of $x_1^{a_1}\cdots x_k^{a_k}$ in the expansion of $$(x_1 + x_2 + \dots + x_k)^{\ell}.$$ The sum of all these coefficients is obtained by substituting $x_1=\dots=x_k=1$.
To eliminate even $a_1$, we can consider the expansion of $$\frac{1}{2}(x_1 + x_2 + \dots + x_k)^{\ell} - \frac{1}{2}(-x_1 + x_2 + \dots + x_k)^{\ell}.$$
Continuing this way, we eventually get $$s(\ell,k) = \frac{1}{2^k} \sum_{t_1,\dots,t_k=0}^1 (-1)^{t_1+\dots+t_k} ((-1)^{t_1}+\cdots+(-1)^{t_k})^{\ell}$$ $$=\frac{1}{2^k} \sum_{z=0}^k \binom{k}{z} (-1)^z (k-2z)^{\ell}.$$
P.S. This formula resembles one for Stirling number of the second kind (formula (10) at MathWorld) but not quite.