As most of you know there is a classical question in elementary combinatorics such that if $a×b×c$=180 , then how many possible positive integer solution are there for the equation ?

The solution is easy such that $180=2^2×3^2×5^1$ and so , for $a=2^{x_1}×3^{y_1}×5^{z_1}$ , $b=2^{x_2}×3^{y_2}×5^{z_2}$ , $c=2^{x_3}×3^{y_3}×5^{z_3}$ .

Then: $x_1+x_2+x_3=2$ where $x_i≥0$ , and $y_1+y_2+y_3=2$ where $y_i≥0$ and $z_1+z_2+z_3=1$ where $z_i≥0$.

So , $C(4,2)×C(4,2)×C(3,1)=108$.

Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $a×b×c<180$ instead of $a×b×c=180$

After , i started to think about it. Firstly , i thought that if i can calculute the possible solutions for $x_1+x_2+x_3<2$ where $x_i≥0$ , and $y_1+y_2+y_3<2$ where $y_i≥0$ and $z_1+z_2+z_3<1$ where $z_i≥0$ , then i can find the solution.However , there is a problem such that when i calculate the solution , i do not include the prime numbers and their multiplicites which is in $180$.

For example , my solution does not contain $1×1×179<180$

My question is that how can we solve these types of question . Is there any TRICK for include all possible ways ? Moreover ,this question can be generalized for $a×b×c≤180$ , then what would happen for it ?