Relatively prime factors of $24500$ 
Let $N=24500$, then find the number of ways by which $N$ can be resolved into two coprime factors?

My tries:
$N=24500=2^2\cdot 5^3\cdot 7^2$, for co prime no those two factors of $24500$ should share something in common, so the two factors should be such that either one of the two should use all of $2's,5's,7's$, hence picking up any two from $2^2\cdot 5^3\cdot 7^2\ \ ,\ \# ={3\choose {2}}=3$, then multiplying them together to get new number, for example $2^2\cdot 7^2=196$, then we are left with $5^3$, hence one such possible resolution should be $196\cdot 5^3$. So, ${3\choose {2}}=3$ should count all answer, but book says answer should be $4$.
Please help.

For integer $a,b,c>0$
$$2^a\cdot5^b\cdot7^c=(1\cdot2^a)(1\cdot5^b)(1\cdot7^c)$$
$=((1\cdot2^a5^b)$ or $(2^a\cdot5^b))(1\cdot7^c)$
$=(1\cdot2^a5^b7^c)$ or $(7^c\cdot2^a5^b)$ or $(2^a7^c,5^b)$ or $(2^a,5^b7^c)$
