If $n_{4} \geq 5$, then $n_{4}!$ contains the prime factors 2 and 5. Therefore,

$$z=n_{1}!n_{2}!n_{3}!n_{4}!-32 \equiv 0-32 \equiv 8(\bmod 10)$$

However, $(10 a+b)^{2} \equiv b^{2}(\bmod 10)$ holds for $a ; b \in N$. If $b=0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9$, then $b^{2} \equiv$ $0 ; 1 ; 4 ; 9 ; 6 ; 5 ; 6 ; 9 ; 4 ; 1(\bmod 10)$.

In no case does $b^{2} \equiv 8(\bmod 10)$ hold. Therefore, $n_{4} \leq 4$. Thus, initially, only the following five quadruples are possible:

$$(0 ; 1 ; 2 ; 3),(0 ; 1 ; 2 ; 4),(0 ; 1 ; 3 ; 4),(0 ; 2 ; 3 ; 4),(1 ; 2 ; 3 ; 4)$$

Testing reveals that the first and third quadruples are invalid. Therefore, the solution set contains three elements: $(0 ; 1 ; 2 ; 4),(0 ; 2 ; 3 ; 4)$, and $(1 ; 2 ; 3 ; 4)$.