4. Given positive integers $a, b, c$ are all distinct, and satisfy $a b c=72^{3}$. Then the number of ordered positive integer triples $(a, b, c)$ is $\qquad$.

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4. 1482 .

Given $72^{3}=2^{9} \times 3^{6}$, let
$a=2^{\alpha_{1}} \times 3^{\beta_{1}}, b=2^{\alpha_{2}} \times 3^{\beta_{2}}, c=2^{\alpha_{3}} \times 3^{\beta_{3}}$.
Then $\alpha_{1}+\alpha_{2}+\alpha_{3}=9, \beta_{1}+\beta_{2}+\beta_{3}=6$, where $\alpha_{i} 、 \beta_{i}(1 \leqslant i \leqslant 3)$ are non-negative integers.
Thus, there are $\mathrm{C}_{9+3-1}^{2} \mathrm{C}_{6+3-1}^{2}=1540$ sets of solutions.
If $a 、 b 、 c$ are all the same, there is only one set of solutions: $a=b=c=72$.

If $a 、 b 、 c$ have only two numbers the same, assume $\alpha_{1}=\alpha_{2}, \beta_{1}=\beta_{2}$, then
$$
2 \alpha_{1}+\alpha_{3}=9,2 \beta_{1}+\beta_{3}=6 \text {, }
$$

Excluding the case where all three numbers are the same, there are $5 \times 4-1=19$ sets of solutions. Considering the order of $a 、 b 、 c$, when only two numbers are the same, the number of ordered tuples $(a, b, c)$ is $19 \times 3=57$.
Therefore, the total number of solutions is $1540-57-1=1482$.