Example 12: Balls are drawn one by one from a bag containing $n$ white balls and $n$ black balls, such that after each draw, the number of black balls remaining in the bag is not less than the number of white balls. Assuming balls of the same color are indistinguishable, find the number of all possible ways to draw the $2n$ balls.
Solution: Find the corresponding "target set". The initial state of drawing balls corresponds to the origin of coordinates. Drawing a white ball corresponds to drawing a diagonal of a unit square with a slope of 1 on the coordinate plane to the right; drawing a black ball corresponds to drawing a diagonal of a unit square with a slope of -1 to the right. During the drawing process, note that the starting point of the next line is the endpoint of the previous one. When all $2n$ balls are drawn, the corresponding broken line extends $2n$ units to the right, but the endpoint is still on the $x$-axis, because the entire broken line consists of $n$ segments rising and $n$ segments falling, so the overall position has neither risen nor fallen, and the endpoint coordinates are $(2n, 0)$.

Furthermore, it is required that the number of white balls drawn is no less than the number of black balls drawn, so the entire broken line must not cross the $x$-axis and intersect with $y=-1$. This means that the methods of drawing balls that meet the requirements correspond one-to-one with the broken lines connecting $O$ and $P$ that do not intersect with $y=-1$. The broken lines connecting $O$ and $P$ are actually the shortest paths from $O$ to $P$ on the chessboard street in Example 3, so the number of broken lines connecting $O$ and $P$ is $C_{2n}^{n}$ (because the rectangle $OMPN$ is $n \times n$), as shown in Figure 30-5.

The broken lines connecting $O$ and $P$ that intersect with $y=-1$ correspond one-to-one with the broken lines connecting $Q(0, -2)$ and $P(2n, 0)$ (which is easy to know by symmetry), and the number of broken lines between $Q(0, -2)$ and $P(2n, 0)$ is $C_{2n}^{n+1}$ (because the rectangle $QRPL$ is $(n+1) \times (n-1)$). Therefore, the number of broken lines connecting $O$ and $P$ that do not intersect with $y=-1$ is $C_{2n}^{n} - C_{2n}^{n+1}$, and thus the number of different ways to draw balls that meet the requirements is $C_{2n}^{n} - C_{2n}^{n+1}$.