Multinomial?

Consider $3z+2y+x = 100$ . We note that for a fixed $z$ and $y$ , there is only one $x$ . Therefore we only need to consider $n(z)$ , the number of solutions of $y$ for a fixed $z$ . And $n(z)$ is equal to the largest possible $y$ for the fixed $z$ . That is $n(z) = \max (y(z)) = \dfrac {100-3z-\min (x)}2$ . Since the $RHS=100$ is even, this implies that when $z$ is odd, $x$ is odd and when $z$ is even, $x$ is even.

Therefore, $\min(x) = \begin{cases} 1 & \text{when }z \text{ is odd.} \\ 2 & \text{ when } z \text{ is even.} \end{cases} \implies n(z) = \begin{cases} \dfrac {99-3z}2 & \text{when }z \text{ is odd.} \\ \dfrac {98-3z}2 & \text{ when } z \text{ is even.} \end{cases}$

Since $z \in [1,32]$ the number of positive integer solutions to the equation is:

$\begin{aligned} \sum_{z=1}^{32} n(z) & = \sum_{k=1}^{16} \left(\frac {99 - 3(2k-1)}2 + \frac {98-3(2k)}2\right) \\ & = \sum_{k=1}^{16} (100 - 6k) = 1600 - \frac {6(16)(17)}2 = \boxed{784} \end{aligned}$

The code below returns the value of 784

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#prints the number of solutions
n = 0
for x in range(100):
    for y in range(50):
        for z in range(33):
            if ((x+1) + 2*(y+1) + 3*(z+1) == 100) :
#remove comment from line below to enumerate all solutions
#                print("Solution: {} {} {}".format(x+1,y+1,z+1))
                n += 1
print("{} solutions".format(n)) 

Output: 784 solutions