The sum of proper divisors of 72 (1 and 72 excluded) is i. 195 ii.122 iii.194 iv. None of these
I have already solved it by adding the divisors (which was easy to do and the only approach I could think of), also the question was mentioned in a book whose only concern is Permutation and Combination so I am wondering if there is some another approach to this problem.
$72=2^3\cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.
That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.
... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.