As n is large, the product n*(n-1)*(n-2) will go out of the range(overflow) and it will return a value different from what is expected. So we consider a shorter product n*(n-1).
n*(n-1) is always an even number. So the subexpression “n * (n-1) / 2 ” in option B would always produce an integer, which means no precision loss in this subexpression. And when we consider `n*(n-1)/2*(n-2)`, it will always give a number which is a multiple of 3.