Example 4: Arrange 5 white stones and 10 black stones in a horizontal row, such that the right neighbor of each white stone must be a black stone. How many arrangements are there?
(1996, Japan Mathematical Olympiad Preliminary)
Let the number of black stones between the $i$-th white stone and the $(i+1)$-th white stone be $x_{i+1}(i=1,2,3,4)$, with $x_{1}$ black stones at the far left and $x_{6}$ black stones at the far right, then
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=10,
$$

where $x_{1} \geqslant 0, x_{k} \geqslant 1, k=2,3,4,5,6$.
$$
\begin{array}{l}
\text { Let } t_{k}=x_{k}-1, k=2,3,4,5,6, \text { then } \\
x_{1}+t_{2}+t_{3}+t_{4}+t_{5}+t_{6}=10,
\end{array}
$$

where $x_{1} \geqslant 0, t_{k} \geqslant 0, k=2,3,4,5,6$.
By the theorem, the equation has $C_{10}^{5}=252$ solutions, meaning there are 252 arrangements.