If $n=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$ 
If $N=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$

My solution:

$$n^2=2^{20}\times 3^{10}$$
Factors of $n^2=\left(20+1\right)(10+1)=21\times 11=231$ 
Factors which are greater than or equal to $n$ can be found as we have choices for power of 2 $\in [10,11,\dots 20]$ and that of 3 $\in[5,6,\dots10]$ 
We get the factors of $n^2$ which are greater than or equal to $n$ as $11\times 6=66$

Factors of $n$ which are less than $n=(10+1)(5+1)-1=65$ ($-1$ as $n$ has been counted in the above condition)

$\therefore$ factors which satisfy the condition are $231-66-65=100$

But the answer is $50$. What is the correct way to solve it and what is the error in my solution?

Let's start with all numbers fitting the desired criteria that have no factors of 2, and increment the power on 2 as we go, and record how many values we find:
$3^6,...,3^{10}$ (5)
$2\times3^6,...,2\times3^{10}$ (5)
$2^2\times3^6,...,2^2\times3^{10}$ (5)
$2^3\times3^6,...,2^3\times3^{9}$ (4)
.
.
.
Notice that higher powers on one base may limit the highest power on the other. List out all possibilities this way and you will get the correct answer.
