Let the notation $|X|$ represent the number of elements in the set $X$. Suppose $|X|=n\left(n \in \mathbf{N}_{+}\right)$, and all subsets of $X$ are $\left\{A_{1}, A_{2}, \cdots, A_{m}\right\}\left(m=2^{n}\right)$. It is known that the sum of the number of elements in the intersections of these subsets is
$$
S=\sum_{i=1}^{m}\left(\sum_{j=1}^{m}\left|A_{i} \cap A_{j}\right|\right)=n \cdot 2^{2(n-1)} .
$$

Try to find the sum
$$
T=\sum_{i=1}^{m-1}\left(\sum_{j=i+1}^{m}\left|A_{i} \cap A_{j}\right|\right) .
$$
(2002, Zhejiang Province Mathematical Summer Camp)
Consider the following $m \times m$ matrix
$$
\left(\begin{array}{cccc}
\left|A_{1} \cap A_{1}\right| & \left|A_{1} \cap A_{2}\right| & \cdots & \left|A_{1} \cap A_{m}\right| \\
\left|A_{2} \cap A_{1}\right| & \left|A_{2} \cap A_{2}\right| & \cdots & \left|A_{2} \cap A_{m}\right| \\
\vdots & \vdots & \vdots & \vdots \\
\left|A_{m} \cap A_{1}\right| & \left|A_{m} \cap A_{2}\right| & \cdots & \left|A_{m} \cap A_{m}\right|
\end{array}\right)
$$

Let the diagonal from the top left to the bottom right be denoted as $l$.
Since the intersection operation of sets is commutative, the matrix is symmetric about $l$. Therefore, the sum of the numbers on $l$ is
$$
R=\sum_{i=1}^{m}\left|A_{i} \cap A_{i}\right|=\sum_{i=1}^{m}\left|A_{i}\right| .
$$

Then $T=\frac{1}{2}(S-R)$.
It is easy to see that $R=\mathrm{C}_{n}^{1}+2 \mathrm{C}_{n}^{2}+\cdots+n \mathrm{C}_{n}^{n}=n \cdot 2^{n-1}$.
Thus, $T=\frac{1}{2}(S-R)=\frac{1}{2}\left[n \cdot 2^{\left.\alpha_{n-1}\right)}-n \cdot 2^{n-1}\right]$
$$
=\frac{n}{2}\left(4^{n-1}-2^{n-1}\right) \text {. }
$$