9. From the set $\{1,2,3, \cdots, 1000\}$, randomly and without replacement, take 3 numbers $a_{1}, a_{2}, a_{3}$, and from the remaining 997 numbers in the set, randomly and without replacement, take another 3 numbers $b_{1}, b_{2}, b_{3}$. Let the probability of the following event be $p$: a rectangular brick of dimensions $a_{1} \times a_{2} \times a_{3}$ can fit into a rectangular box of dimensions $b_{1} \times b_{2} \times b_{3}$ after appropriate rotation, with the edges of the brick parallel to the edges of the box. If $p$ is written as a reduced fraction, what is the sum of the numerator and the denominator?
9. Solution: A rectangular brick of size $a_{1} \times a_{2} \times a_{3}$ can fit into a rectangular box of size $b_{1} \times b_{2} \times b_{3}$ if $a_{1}<b_{1}, a_{2}<b_{2}, a_{3}<b_{3}$. Given $c_{0}<c_{6}$, $c_{1}$ must be one dimension of the brick, and $c_{6}$ must be one dimension of the box. Here, we refer to the length, width, and height of a rectangular prism as its three dimensions.

From the four numbers $c_{2}, c_{3}, c_{4}, c_{5}$, any two (excluding the pair $\left(c_{4}, c_{5}\right)$) can be chosen as the other two dimensions of the brick, and the remaining two numbers can be the other two dimensions of the box. Therefore, there are $C_{4}^{2}-1=5$ (ways). Thus, $P=\frac{C_{1000}^{6} \times 5}{C_{1000}^{3} \cdot C_{997}^{3}}=\frac{1}{4}$.
Therefore, the required probability is $\frac{1}{4}$, and the sum of the numerator and denominator of $P$ is 5.