Seemingly simple combinatorial problem Count all $n$-length strings of digits $0, 1,\dots, m$ that have an equal number of $0$'s and $1$'s. Is there a closed form expression?

Assuming we want the number of strings of $n$ characters over the alphabet $\{0,\ldots,m\}$ with the same number of "zeroes" and "ones" we just have to compute:
$$\sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{2j}\binom{2j}{j}(m-1)^{n-2j}.$$
Reason: set $j$ as the number of zeroes, then choose the $2j$ over $n$ places where putting all the zeroes and ones, then choose where to put the zeroes over such $2j$ positions, the choose how to fill the remaining part of the string with characters from $\{2,\ldots,m\}$ ($(m-1)$ possible choices).
Expanding $(m-1)^{n-2j}$ with the binomial theorem we get:
$$\sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{2j}\binom{2j}{j}\sum_{k=0}^{n-2j}\binom{n-2j}{k}m^k (-1)^{n-2j-k},$$
and the balanced strings are generated by the grammar:
$$ S \to 0S1,1S0,SS,\varepsilon.$$
