8. From the numbers $1,2, \cdots, n$, if two different numbers are chosen and multiplied, the arithmetic mean of the products is $\qquad$
8. $\frac{(n+1)(3 n+2)}{12}$.

The arithmetic mean of the products is equal to
$$
\begin{array}{l}
\frac{1}{\mathrm{C}_{n}^{2}} \sum_{i<i<j<n} i j=\frac{1}{n(n-1)}\left[\left(\sum_{i=1}^{n} i\right)^{2}-\sum_{i=1}^{n} i^{2}\right] \\
=\frac{(n+1)(3 n+2)}{12} \text {. } \\
\end{array}
$$