Find the coefficient of $x^{24}$ in the binomial equation

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$ds{bracks{cdots}}$ "inside the sums" is an Iverson Bracket.

I suggest don't expand the terms. There are three expressions in multiplication which give power of $x$ as multiple of $1$, $2$ and $3$ respectively. Let the exponent of $x$ from each expression be $n_1, , n_2$ and $n_3$ respectively where $n_iinmathbb{W}$.

So, you've to find no. of non-negative integral solutions of $n_1+2n_2+3n_3=24$.

If you think about the product as $$frac{1}{(1-x)(1-x^2)(1-x^3)} = (1+x+x^2+cdots)(1+x^2+x^4+cdots)(1+x^3+x^6+cdots),$$ it becomes clear that the coefficient of $x^{24}$ is given by the number of ways to choose nonnegative integers $a,b,c$ such that $a + 2b + 3c = 24.$ Fortunately, these are not too difficult to enumerate since $24$ is small. The best way to proceed is to first choose $c$; i.e., for each $c in {0, 1, 2, ldots, 8}$, solve $$a + 2b = 24 - 3c = 3(8-c).$$ This in turn requires $$a = 3(8-c) - 2b ge 0,$$ so $b in {0, 1, ldots, 12 - lceil tfrac{3}{2}c rceil}$. So that means there are $$sum_{c=0}^8 12 - leftlceil frac{3}{2} c rightrceil + 1 = 13(9) - (0 + 2 + 3 + 5 + 6 + 8 + 9 + 11 + 12) = 61$$ such solutions, and this is the desired coefficient.