For the positive integer $n$, it is defined that $n!=1 \times 2 \times \cdots \times n$. Then, among all the divisors of the product $1! \times 2! \times \cdots \times 9!$, the number of divisors that are perfect squares is $\qquad$.

7.672.

Given $1!\times 2!\times \cdots \times 9!=2^{30} \times 3^{13} \times 5^{5} \times 7^{3}$, then the divisors of its perfect square must be in the form of $2^{2 a} \times 3^{2 b} \times$ $5^{2 c} \times 7^{2 d}$, where $a, b, c, d$ are non-negative integers, and $0 \leqslant a \leqslant 15,0 \leqslant b \leqslant 6,0 \leqslant c \leqslant 2,0 \leqslant d \leqslant 1$. Therefore, such divisors total $16 \times 7 \times 3 \times 2=672$.