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

Since we are looking for a set, $\{2310,1,1\}$ is not an answer - that is, not all the prime factors are gathered in the same factor.
Then, to avoid double-counting, we can say, for example, that $11$ always divides $a$ and the largest prime factor $q$ that does not divide $a$, divides $b$.
So for $q=7$, we can distribute the smaller primes arbitrarily among the three factors, giving $3^3$ options. Similarly for $q=\{5,3,2\}$ there are $\{3^2, 3^1, 3^0\}$ options giving a total of $27+9+3+1=\fbox{40}$ options altogether.
