For how many three-element sets (order doesn't matter) of positive integers {a,b,c} is it true that abc = 2310 (The positive integers a, b, and c are distinct.)
The prime factorization of 2310 is 2*3*5*7*11, so I split this into cases (if one of the elements is 1, and if none of the elements are 1)
if one of the elements is 1 then we can choose either 1 or 2 of the primes as the first factor, and the rest of the primes as the last factor
$\binom51\binom44+\binom52\binom33=5\cdot1+10\cdot1=15$
if none are 1 then we choose 1 prime as first factor, 1 or 2 for second, and the rest as the third
$\binom51\binom41\binom33+\binom51\binom42\binom22=5\cdot4\cdot1+5\cdot6\cdot1=50$
The answer is then 15 + 50 = 65.
