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 $\big(\sin(x)\big)^k$.



What is this restricted sum of multinomial coefficients?
$\binom{\ell}{a\_1,\dots,a\_k}$ 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](http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html)) but not quite.


