Lets try to prime factorize $10!$ because each if $p^n$ is a factor of a number $x$ for any prime number $p$, then $x$ has at least $p+1$ factors -- $p^0, p^1, p^2, \ldots, p^n.$
$10!=2\times3\times4\times5\times6\times7\times8\times9\times10.$
$=2^8 \times 3^4 \times 5^2 \times 7$
Now, we can count the no. of factors of $10!$ which will be $(8+1) \times (4+1) \times (2+1) \times (1+1) =270.$ For each of the factor $d$ of any number $x$, we can get $x$ by multiplying $d$ by $x/d$ where $x/d$ is another factor of $x$. So, number of ways we can choose $(a,b)$ such that $a\times b = 10!$ will be $\frac{270}{2} = 135.$ In multiplication there is no need to consider $a \times b$ different from $b \times a$, otherwise answer would be $270$.