Example 2 Consider a $19 \times 2015$ grid. A "block" is formed by a $10 \times 10$ grid and a cell to the right of its top-right cell. It is known that the cells of different blocks do not overlap, and the cells of the blocks coincide with the cells of the grid. If blocks cannot be rotated, find the number of ways to place the maximum number of blocks in the grid. ${ }^{[2]}$
(2015, Bulgarian National Team Selection Exam)
Let the number of blocks that can be placed be $n$. Color the top-right corner of each block black. Since each block occupies 10 rows, and the grid has only 19 rows, the black cells must be in different columns. Therefore, we can sort the black cells from left to right. For $i=1,2, \cdots, n$, let the $i$-th black cell be in the $a_{i}$-th row and $b_{i}$-th column. Thus,
$$
\begin{array}{l}
1 \leqslant a_{i} \leqslant 10(i=1,2, \cdots, n), \\
b_{1} \geqslant 11, b_{i+1} \geqslant b_{i}+10(i=1,2, \cdots, n-1),
\end{array}
$$

and when $b_{i+1}=b_{i}+10$, it must be that $a_{i+1} \geqslant a_{i}+1$.
Therefore, when $i+10 \leqslant 2015$, it must be that $b_{i+10} \geqslant b_{i}+101$ (otherwise, it can only be $b_{i+10}=b_{i}+100$, which would imply $a_{i+10} \geqslant a_{i}+10>10$, a contradiction).
If $n \geqslant 200$, then
$$
\begin{array}{l}
b_{200} \geqslant b_{10}+101 \times 19 \\
\geqslant b_{1}+10 \times 9+101 \times 19 \\
=b_{1}+2009>2015,
\end{array}
$$

a contradiction.
Thus, $n \leqslant 199$.
Next, consider the number of ways to place 199 blocks.
Let $c_{i}=b_{i}-10 i(i=1,2, \cdots, 199)$. Then
$$
1 \leqslant c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{199} \leqslant 2015-1990=25,
$$

and when $c_{i+1}=c_{i}$, it must be that
$$
a_{i+1} \geqslant a_{i}+1(i=1,2, \cdots, 198).
$$

Thus, $\left(a_{i}, c_{i}\right)(i=1,2, \cdots, 199)$ can correspond to 199 distinct cells in a $10 \times 25$ grid, such that these cells appear in the order of the 1st column, 2nd column, ..., 25th column from top to bottom. Conversely, for any 199 distinct cells in a $10 \times 25$ grid, we can assign coordinates $\left(a_{i}, c_{i}\right)$ $(i=1,2, \cdots, 199)$ in the above order, uniquely corresponding to a distribution of black cells, and thus uniquely corresponding to a way of placing 199 blocks.

Therefore, the number of ways to place the maximum number of blocks (199 blocks) is $\mathrm{C}_{250}^{199}$.