110 = 2 * 5 * 11

Number of ordered positive triplets = (1 + 3 - 1)C2 * (1 + 3 - 1)C2 * (1 + 3 - 1)C2 = 3 * 3 * 3 = 27

Ordered integer triplets = 4 * 27 = 108 (considering the possibilities (-a, -b, c), (-a, b, -c), (a, -b, -c), (a, b, c))

This includes cases like aab and aaa, which needs to be removed to get the unordered pairs.

a = b = c : no cases possible

a = b or b = c or c = a

As all the prime factors only once, we cannot split it into cases where 2 entries are equal. So the only cases are

(110, 1, 1) & (110, -1, -1) -> which can be arranged in 3!/2! = 3 ways each

All the other triplets can be arranged in 3! ways

So final unordered triplets = (108 - 6)/3! + 6/3 = 102/6 + 2 = 17 + 2 = 19 ways.