Clearly the factors are $2310=2\times 3\times 5 \times 7 \times 11$
My approach was picking one number for $a$, this leaves 4 numbers for b and c. So b can take up 1 or 2 or 3 numbers out of those 4 numbers
So
$$\binom 51 (\binom 41 +\binom 42 +\binom 43)$$
$$=5(2^4-2)$$
Similar, 2 numbers can be picked for $a$, leaving 3 for b and c
Continuing this way gives
$$5(2^4-1) +10 (2^3-2) + 10 (2^2-2) $$which isn’t even close to 40, the answer.
What am I doing wrong?
I know this question already exists on MSE, but unfortunately it doesn’t solve my specific problem.
NOTE: I am using this problem as a reference for my answer The students $S_1, S_2,…S_{10}$ are divided into 3 groups A, B and C
Best Answer
Knowing the prime factors makes things a lot easier. We basically just assign each factor to any of $a$, $b$, or $c$ and call it good. This means that there are a total of $3^5$ possible ways of assigning these factors, a total of $243$ ways.
This, however, overcounts several conditions. Firstly, if we ever end up in a situation where two of the numbers are the same, then we don't get a $3$ element set, so we need to exclude these. These conditions occur if ever the distribution of the prime factors could lead to two equal products. But, by the uniqueness of prime factorization, since all of our primes are distinct this can only happen if two of the numbers were both $1$, meaning that all of the factors ended up in a single number. This happens in exactly $3$ situations, so we remove these three.
We have still overcounted, however, since we do not care about the permutations of $a$, $b$, and $c$. We divide this new total by the total number of permutations that would lead to this collection of numbers, which is a total of $3!$ permutations. In all, we have
$$\frac{3^5-3}{3!}=\frac{243-3}{6}=\frac{240}{6}=40$$
Hope this helps!