Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{1}}7^{w_{1}}11^{t_{1}}$
$b=2^{x_{2}}3^{y_{2}}5^{z_{2}}7^{w_{2}}11^{t_{2}}$
$c=2^{x_{3}}3^{y_{3}}5^{z_{3}}7^{w_{3}}11^{t_{3}}$
where $x_{1}+x_{2}+x_{3}=1$; $x_{1},x_{2},x_{3}$ being non-negative integers.
Number of solutions of this equation clearly $3$ and similarly for other variables as well.
So,number of solutions $=3^5$.
But answer given is $40$ .
Am I misinterpreting the question

I got a combinatorial  solution.
$2310=2 \times 3 \times 5 \times 7 \times 11$
$\dfrac{\dbinom{5}{1}\dbinom{4}{1}\dbinom{3}{3}}{2!}=10\\
\dfrac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2}}{2!}=15\\
\dbinom{5}{3}\dbinom{2}{2}=10\\
\dbinom{5}{4}\dbinom{1}{1}=5$
Total: $40$
