Example 3 Put 1996 identical balls into 10 different boxes, so that the $i$-th box contains at least $i$ balls. How many different ways are there to do this?
Solution: Let the $i$-th box contain $x_{i}\left(x_{i} \geqslant i, i=1\right.$, $2, \cdots, 10)$ balls, then
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{10}=1996 . \\
\text { Let } x_{i}^{\prime}=x_{i}-i, \text { then } \\
x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{10}^{\prime}=1941\left(x_{i}^{\prime} \geqslant 0\right) .
\end{array}
$$

By the theorem, the system of equations has $\mathrm{C}_{1950}^{9}$ solutions.
Therefore, there are $\mathrm{C}_{1950}^{9}$ different ways to place the balls.