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
Best Answer
$3^5=243$ 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 $40$.