Find number of integral solutions of a*b*c*d = 600

The prime factors of $600$, expressed as a multiset, are $\{2,2,2,3,5,5\}$. To get the original answer of $800$ ordered tuples, we partition each distinct prime across $(a,b,c,d)$ with stars & bars to get: $${6 \choose 3}{4 \choose 3}{5\choose 3} = 20\cdot 4\cdot 10 = 800$$

Clearly we can never have all four factors the same. We can have three factors the same only in $2$ distinct cases, $(1,1,1,600)$ and $(2,2,2,75)$, which correspond to $8$ choices in the $800$.

Two factors the same, $(a,a,X,Y)$, can have $a \in \{1,2,5,10\}$. For $a=1$ in given position, there are ${4 \choose 1}{2 \choose 1}{3\choose 1}=24$ options, and similarly $a=(2,5,10)$ have $(12,8,4)$ options. For $a=(1,2)$ we also need to reduce the values by $2$ to remove the overlap with the three-equal-factor case. The total of $44$ options correspond to $44\cdot {4 \choose 2} = 264$ cases in the $800$ but only $22$ in the distinct cases.

The remaining cases, $800-264-8 = 528$, have four distinct values and are a multiple of $4!$ of the distinct cases, giving $528/24 = 22$ more distinct cases, for a total of: $$ 2+22+22 = 46 \text{ distinct cases}$$

You just have to be systematic. We want $a\le b\le c\le d$. We have $5^4=625$, so $a\le4$. If $a=4$, then $bcd=150$. We have $6^3=216$, so $b<6$. Hence $b=5$. So $cd=30$ and hence $c=5,d=6$.

If $a=3$, then $bcd=200$, so $b=4$ or 5. If $b=4$, then $cd=50$, so $c=5,d=10$. If $b=5$, then $cd=40$, so $c=5,d=8$.

If $a=2$, the $bcd=300$, so $b<7$. If $b=2$, then $cd=150$, so $c=2,3,5,6,10$. If $b=3$, then $cd=100$, so $c,d=(4,25),(5,20),(10,10)$. If $b=4$, then $cd=75$, so $c=5,d=15$. If $b=5$, then $cd=60$, so $c,d=(5,12),(6,10)$. If $b=6$, then $cd=50$, which does not give any solutions.

If $a=1$, then $bcd=600$, so $b\le8$. If $b=1$, then $cd=600$, so we can have $c=1,2,3,4,5,6,8,10,12,15,20,24$. If $b=2$, then $cd=300$, so we can have $c=2,3,4,5,6,10,12,15$. If $b=3$, then $cd=200$, so we can have $c=4,5,8,10$. If $b=4$, then $cd=150$, so we can have $c=5,6,10$. If $b=5$, then $cd=120$, so we can have $c=5,6,8,10$. If $b=6$, then $cd=100$, so we can have $c=d=10$. If $b=8$, then $cd=75$, which has no solutions.

Summary: $(a,b,c,d)=$(4,5,5,6),(3,4,5,10),(3,5,5,8),(2,2,2,75),(2,2,3,50),(2,2,5,30),(2,2,6,25),(2,2,10,15),(2,3,4,25),(2,3,5,20),(2,3,10,10),(2,4,5,15),(2,5,5,12),(2,5,6,10),(1,1,1,600),(1,1,2,300),(1,1,3,200),(1,1,4,150),(1,1,5,120),(1,1,6,100),(1,1,8,75),(1,1,10,60),(1,1,12,50),(1,1,15,40),(1,1,20,30),(1,1,24,25),(1,2,2,150),(1,2,3,100),(1,2,4,75),(1,2,5,60),(1,2,6,50),(1,2,10,30),(1,2,12,25),(1,2,15,20),(1,3,4,50),(1,3,5,40),(1,3,8,25),(1,3,10,20),(1,4,5,30),(1,4,6,25),(1,4,10,15),(1,5,5,24),(1,5,6,20),(1,5,8,15),(1,5,10,12),(1,6,10,10). Total 46 solutions.