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
 A: $3^5=243$ is the number of ordered triples $\{a,b,c\}$ such that $abc=2310$. We are asked for the number of sets. So (i) $a,b,c$ must all be different; and (ii) their order is irrelevant.
Condition (i) rules out triples $\{1,1,2310\},\{1,2310,1\},$ and $\{2310,1,1\},$ which leaves 240.
Condition (ii) means we should divide this by $3!=6$, because each set corresponds to six triples. So we are left with $40$.
A: 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: $40$
A: A different approach: You need to get three factors of $2310$ out of your five numbers. There are two ways to do that. Either include a factor of $1$, or don't.
$A$ (Without having $1$ as a factor): 
i) You can have two of the five prime factors as two elements of your set, multiplying the remaining three prime factors together to get the third element. There are $10$ ways to do that.
ii) You can have one of the five prime factors as one element of your set, and then multiply the remaining four prime factors together in pairs, to give the remaining two elements of your set. There are $15$ ways to do that.
$B$ (Having 1 as a factor): 
Have a factor of $1$ as one of the elements of your set, and then multiply your five prime factors together to get the two other elements of your set. There are $2$ ways to do that:
i) Choose three of them, multiply those, make that your second element of your set, and then multiply the remaining two for the third element. There are $10$ ways to do that. 
ii) Choose four of them, multiply those, make that the second element of your set, and then have the remaining one as your third element. There are $5$ ways to do that.
This gives $40$.
A: $5=3+1+1, 1+2+2.$ So making sets without integer $1,$ first possibility, $10$ ways. Second possibility, $15$ ways. If integer $1$ is taken once $, 5=3+2, 4+1.$
First possibility $10$ ways, second possibility $5$ ways. So total $40$ ways.
A: 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.
