Example 2-62 A binary number of $2n$ bits composed of $n$ ones and $n$ zeros, requires that the cumulative number of 1s is greater than the cumulative number of 0s when scanning the first $2n-1$ bits from left to right. Find the number of such numbers. This problem can be reduced to finding the number of paths from point $O$ to point $A$ that only pass through points above the diagonal $OA$, which is equivalent to finding the number of paths from point $O'(0,1)$ to point $A'(n-1, n)$ without crossing the diagonal $OA$. Thus, it is transformed into the problem of the previous example.
According to the result of the previous example, the number of paths from point $O^{\prime}(0,1)$ through point $O^{\prime} A^{\prime}$, and above point $O^{\prime} A^{\prime}$ to reach point $A^{\prime}(n-1, n)$ is
$$
\begin{aligned}
\binom{2 n-2}{n-1}-\binom{2 n-2}{n} & =(2 n-2)!\left[\frac{1}{(n-1)!(n-1)!}-\frac{1}{n!(n-2)!}\right] \\
& =(2 n-2)!\frac{n-n+1}{n!(n-1)!} \\
& =\frac{1}{n}\binom{2 n-2}{n-1}=h_{n+1}
\end{aligned}
$$