Example 23 (2003 Anhui Provincial Competition Question) Given that $x$, $y$, and $z$ are all positive integers, the equation $x+y+z=15$ has $\qquad$ groups of solutions.
Solution: Fill in 91. Reason: Since we are looking for positive integer solutions, it is clear that $1 \leqslant x, y, z \leqslant 13$.
Therefore, the original problem can be transformed into: writing 15 as the sum of 15 ones, i.e., $1+1+\cdots+1=15$, where any 2 of the 14 plus signs are chosen, and the 1s separated by these two plus signs are combined into one number to get the solution of the equation. Thus, the number of solutions is $C_{14}^{2}=91$.