Mongolian Mathematical Olympiad

Let $a$, $b$, $c$, $d$, $e$, $f$ be not necessarily distinct divisors of $210$. First we will find number of $6$-permutations $(a,b,c,d,e,f)$ that satisfy the condition $abcdef>210^3$. $$\text{Since }abcdef>210^3\iff \frac{210}{a}\cdot \frac{210}{b}\cdot \frac{210}{c}\cdot \frac{210}{d}\cdot \frac{210}{e}\cdot \frac{210}{f}<210^3.$$ Number of $6$-permutations which satisfy the condition $abcdef>210^3$ equals to number of $6$-permutations which satisfy the condition $abcdef<210^3$. Number of $6$-permutations with $abcdef=2^3\cdot 3^3\cdot 5^3\cdot 7^3$ equals to $((\genfrac{}{}{0ex}{}{6}{3}){)}^4$. Therefore number of $6$-permutations $(a,b,c,d,e,f)$ which satisfy the condition $abcdef>210^3$ equals to $\frac{(16{)}^6-(\left(\genfrac{}{}{0ex}{}{6}{3}\right){)}^4}{2}$.

Now we apply this result to the given problem. Setting $f=210$ we get $abcde>210^2$ and thus desired number is $\frac{1}{6}\cdot \frac{(16{)}^6-(\left(\genfrac{}{}{0ex}{}{6}{3}\right){)}^4}{2}$.