Problematic Combinations

First of all, we need to find X:

The prime factorisation of 2160 is $2^4 \times 3^3 \times 5$ . All numbers are uniquely defined by their prime factorisation so it is easier to consider a, b and c as the product of their prime factors because we know that the prime factors of a, b and c combine (multiply together) to form the prime factorisation of 2160. So we can think of this problem as the number of ways to split $2^4 \times 3^3 \times 5$ between three integers. We have four 2s to split between the three integers a, b and c. There isn't necessarily a 2 in all of them so using stars and bars, we have 2 partitions and 4 stars (the twos) so there are 6C2 ways to position the partitions in the list so 15 ways to split up the twos. Similarly, there are 5C2 ways to split up the threes and 3 places to put the five. Notice that it is okay if, for example, there are no twos, threes or fives in the prime factorisation of $a$ because in that situation $a$ is 1. So:

X = 6C2 x 5C2 x 3 = 450

Now to find Y:

We need to split 2160 between three integers. a, b and c are all positive so, using the idea of stars and bars, there are 2160 stars, two partitions and 2159 gaps so there are 2159C2 combinations. So:

Y = 2159C2 = 2329561

Therefore X + Y = 2329561 + 450 = $\boxed{2330011}$ .