Highest common factor of divisors of a number 
Given that a number is N=2910600 . Find total number of ways in which given number can be split into two factors such that their highest common factor is a prime number .

my attempt
$N=2^3 \cdot 3^3 \cdot 5^2 \cdot 7^2 \cdot 11$
Now, one of the number would have only the first power of a prime number and the second number would have the remaining powers of the chosen prime number. The other prime numbers could be multiplied either with factor 1'or factor 2.
Example 
Factor 1: $2^1$  
Factor 2 : $2^2 \cdot 3^3 \cdot 5^2 \cdot 7^2 \cdot 11$
Or 
Factor 1: $2^1 \cdot 3^3 \cdot 5^2 \cdot 7^2 \cdot 11 $
Factor 2: $2^2$ 
This makes the answer $2\cdot 5= $10
Is my way correct or is there an alternate solution ?

Suppose that $n=ab$ with $\gcd(a,b)=p$ for some prime $p$. So $a=a'p$ and $b=b'p$ for some natural numbers $a'$ and $b'$. Hence $N=a'b'p^2$ with $\gcd(a',b')=1$.
So in order to construct all possible pairs $(a,b)$, we pick a prime $p$ and divide $N$ by $p^2$. Then we split the quotient in two coprime factors, $(a',b')$.
Now, $N=2^3\cdot 3^3\cdot 5^2\cdot 7^2\cdot 11$
So we have $4$ choices for $p$, namely $2$, $3$, $5$ and $7$. If we choose $5$ or $7$, the quotient has $4$ unique prime factors. For each of these we must decide whether they will be a factor of $a'$ or not (meaning they'll be a factor of $b$)'. This gives $2^4$ possible combinations, but some of these will be the same, with $a'$ and $b'$ switched, so we'll divide by $2$ and there are $8$ possible combinations.
If we choose $2$ or $3$ for $p$, there are $5$ unique primes left, so there are $\frac12\cdot 2^5=16$ possible combinations.
This gives a total number of possible ways of:
$$8+8+16+16=48$$ 
different ways.
