Example 2-61 $n$ ones and $n$ zeros form a $2n$-bit binary number. It is required that when scanning from left to right, the cumulative number of 1s is not less than the cumulative number of 0s. Try to find how many numbers satisfy this condition.
Below are two solutions.

Solution 1: Let $p_{2n}$ be the number of such numbers. The number of ways to fill $2n$ positions with $n$ 1s is $\binom{2n}{n}$. The remaining $n$ positions are automatically filled with 0s. Subtracting the number of invalid schemes from $\binom{2n}{n}$ gives the desired result. An invalid number is one where, when scanned from left to right, the cumulative count of 0s exceeds the cumulative count of 1s.

The characteristic of an invalid number is that when scanned from left to right, there must be a certain odd position $2m+1$ where the cumulative count of 0s first exceeds the cumulative count of 1s, with $m+1$ 0s and $m$ 1s. In the subsequent $2(n-m)-1$ positions, there are $n-m$ 1s and $n-m-1$ 0s. If the 0s and 1s in this part of $2(n-m)-1$ positions are swapped, making it consist of $n-m$ 0s and $n-m-1$ 1s, the result is a $2n$-bit number composed of $n+1$ 0s and $n-1$ 1s. That is, an invalid number corresponds to a permutation of $n+1$ 0s and $n-1$ 1s.

Conversely, any $2n$-bit number composed of $n+1$ 0s and $n-1$ 1s, since the number of 0s is two more, and $2n$ is even, must have a certain odd position where the cumulative count of 0s exceeds the cumulative count of 1s. Similarly, in the subsequent part, swapping 0s and 1s makes it a $2n$-bit number composed of $n$ 0s and $n$ 1s. That is, a $2n$-bit number composed of $n+1$ 0s and $n-1$ 1s must correspond to an invalid number.

The above method establishes a one-to-one correspondence between $2n$-bit numbers composed of $n+1$ 0s and $n-1$ 1s, and $2n$-bit numbers composed of $n$ 0s and $n$ 1s where the cumulative count of 0s exceeds the cumulative count of 1s when scanned from left to right.

For example, $1010\dot{0}101$ is an 8-bit binary number composed of 4 0s and 4 1s, but when scanned from left to right, at the 5th position (marked with *), the cumulative count of 0s 3 exceeds the cumulative count of 1s 2, corresponding to a 10100010 composed of 3 1s and 5 0s.
Conversely, $10100\dot{0}010$ corresponds to 10100101.
Thus, invalid $2n$-bit numbers correspond one-to-one with permutations of $n+1$ 0s and $n-1$ 1s, hence
$$
\begin{aligned}
p_{2n} & =\binom{2n}{n}-\binom{2n}{n+1}=(2n)!\left[\frac{1}{n!n!}-\frac{1}{(n-1)!(n+1)!}\right] \\
& =(2n)!\left[\frac{n+1}{(n+1)!n!}-\frac{n}{(n+1)!n!}\right]=\frac{(2n)!}{(n+1)!n!} \\
& =\frac{1}{n+1}\binom{2n}{n}
\end{aligned}
$$

Solution 2: This problem can be one-to-one corresponded to the path from the origin $(0,0)$ to the point $(n, n)$, requiring that the points $(a, b)$ passed through satisfy the relation $a \leqslant b$.

The corresponding method is to start from $(0,0)$ and scan the $2n$-bit number from left to right. If a 1 is encountered, move one step in the positive $y$-direction; if a 0 is encountered, move one step in the positive $x$-direction. Since there are $n$ 0s and $n$ 1s, it corresponds to a path from $(0,0)$ to $(n, n)$. Since the cumulative count of 1s must not be less than the cumulative count of 0s, it can pass through points on the diagonal $OA$ but must not cross the diagonal. Conversely, a path satisfying this condition corresponds to a $2n$-bit binary number that meets the requirements.

The problem is transformed into finding the number of paths from $(0,0)$ to $(n, n)$ that pass through points on or above the diagonal $OA$.

The discussion of the number of paths from one point to another is covered in Chapter 1. As shown in Figure 2-23, the path from $O$ to $A$ passing through points on or above the diagonal $OA$ corresponds to a path from $O'$ to $A'$ passing through points on or above the diagonal $O'A'$. This is obvious.

The path from $O'$ to $A'$ passing through points on the diagonal $OA$ is the path from $O'$ to $A'$ crossing the diagonal $O'A'$. Hence, it corresponds to a path from $O$ to $A$ crossing the diagonal $OA$.

Therefore, the number of paths from $O$ to $A$ passing through points on or above the diagonal $OA$ is equal to the number of all paths from $O'$ to $A'$ minus the number of paths from $O'$ to $A'$ passing through points on the diagonal $OA$. That is,
$$
\begin{aligned}
& p_{2n}=\binom{2n}{n}-\binom{2n}{n+1}=(2n)!\left[\frac{1}{n!n!}-\frac{1}{(n+1)!(n-1)!}\right] \\
= & \frac{1}{n+1}\binom{2n}{n}=h_{n+2}
\end{aligned}
$$