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 $abcdef > 210^3$.
$$\text{Since } abcdef > 210^3 \Leftrightarrow \frac{210}{a} \cdot \frac{210}{b} \cdot \frac{210}{c} \cdot \frac{210}{d} \cdot \frac{210}{e} \cdot \frac{210}{f} < 210^3.$$
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 $(\binom{6}{3})^4$. Therefore number of $6$-permutations $(a, b, c, d, e, f)$ which satisfy the condition $abcdef > 210^3$ equals to $\frac{(16)^6 - (\binom{6}{3})^4}{2}$.

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