Question

The number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310,$ is

• A. 40
• B. 30
• C. 25
• D. 15

Answer 1

Answer: (A)

$N=2310=2\cdot 3\cdot 5\cdot 7\cdot 11$
We need to either partition these $5$ prime numbers in groups of $2$ with the other third number $1$ or groups of $3$ .
Now
$5=1+4=2+3$ and $5=1+1+3=1+2+2$
Hence, number of ways
$=\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{1!1!2!3!}+\frac{5!}{1!2!2!2!}$
$=5+10+10+15=40$