In how many ways can a number be written as a product of two different factors? 
In how many ways can a number be written as a product of two different
  factors?

MY APPROACH :
FOR EXAMPLE:Let the number be $8100$=$2^23^45^2$.
Number of divisors = $(2+1)(4+1)(2+1) = 45$
The divisors can be written as product of two numbers like.

$1 \times 8100$
$2 \times 4050$
$3 \times 2700$
$4 \times 2025$
$5 \times 1620$
.
.
.
$8100 \times 1$

Due to repetition a top-bottom symmetry can be observed.Hence the actual number of ways of representing as a product should be 

$$\frac{45+1}{2}=\frac{46}{2}=23$$

Is my approach correct?Can someone please verify whether my argument that the final answer should be $\frac{45+1}{2}$ is valid?
EDIT:As @mathlove pointed out we need to subtract 1 as $90*90$ is not allowed as the question asks for different factors.Final result $\frac{45-1}{2}=22$.
 A: You are almost right. As already observed in the comments,  note that, calling $k_1, k_2, k_3... $ the exponents of each prime number in the factorization,  if your final product $(k_1+1)(k_2+1)(k_3+1)... $ is odd, then this means that all exponents are even and so the number is a perfect square. In this case, to find the total nunber of ways to express the number as a product of  two factors (regardless of whether these two factors are equal or different) you would have to add $ 1$ to your final product and divide to 2, since the symmetric distributions of the prime numbers in two factors $a \cdot b $ include one where $a=b $. For example, for $36= 2^2 \cdot 3^2$, we have $3 \cdot 3=9$ "total" divisors  ($1, 2, 3, 4, 6, 9, 12, 18, 36$) but $5$ ways to express $36$ as a product of two factors ($36 \cdot 1, 18 \cdot 2, 12 \cdot 3, 9 \cdot 4, 6 \cdot 6$).
So if the question talks about two "different" factors, you have to subtract 1 and divide to 2 (in this example we have to exclude the product  $6 \cdot 6$).
In contrast, if the final product is even, then the number is not a perfect square and so you only have to divide to 2. 
A: Yes, you are globally correct.


*

*Let $n \in N , n = \sum p_i^{w_i}$ , the prime decomposition of $n$.

*For the 1st factor, by choosing all the combinations of the powers of the primes from $0$ to their effective values, we get the number of ways to write $n$ with two factors : $S_1 = \prod (w_i+1)$. The second number is merely unique , it is the quotient of $n$ by the 1st found.

*Now, we want to count only the pairs of different factors $S_2$, let's check if $n$ is a square, or if all the $w_i$ are even. In this case, decrease the previous result by $1$

*Finally, it we consider that each solution $(a,b)$ is the same than $(b,a)$, then $S_3 = \frac{S_2}2$. We can do it since we have removed the only possible same factor pair.
