29. Let $p$ be a prime, $a, b \in \mathbf{N}^{*}$, satisfying: $p>a>b>1$. Find the largest integer $c$, such that for all $(p, a, b)$ satisfying the conditions, we have
$$p^{c} \mid\left(\mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}\right)$$
29. When taking $p=5, a=3, b=2$, we should have $5^{c} \mid 3000$, so $c \leqslant 3$. Below, we prove that for any $p, a, b$ satisfying the conditions, we have $p^{3} \mid \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}$.

In fact, notice that
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) \cdot \cdots \cdot((a-b) p+1)}{(b p)!}-\frac{a!}{b!(a-b)!} \\
= & \frac{a(a-1) \cdot \cdots \cdot(a-b+1) \prod_{k=a-b}^{a-1}(k p+1)(k p+2) \cdots \cdots \cdot(k p+(p-1))}{b!\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdots \cdots(k p+(p-1))} \\
& -\frac{a!}{b!(a-b)!} \\
= & \frac{a!}{b!(a-b)!} \cdot \frac{1}{\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))}\left\{\prod_{k=a-b}^{a-1}(k p+1)\right. \\
& \left.\cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))-\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))\right\},
\end{aligned}$$

Thus, we only need to prove that $p^{3} \mid A$, where
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1)) \\
& -\prod_{k=0}^{b-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))
\end{aligned}$$

For this, we set
$$\begin{aligned}
f(x) & =(x+1)(x+2) \cdot \cdots \cdot(x+(p-1)) \\
& =x^{p-1}+\alpha_{p-2} x^{p-2}+\cdots+\alpha_{1} x+(p-1)!
\end{aligned}$$

Then, by the conclusion of Example 1 in Section 2.4, we know that $p^{2} \mid \alpha_{1}$. Therefore,
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1} f(k p)-\prod_{k=0}^{b-1} f(k p) \\
\equiv & ((p-1)!)^{b-1} \sum_{k=a-b}^{a-1} \alpha_{1} k p+((p-1)!)^{b} \\
& -((p-1)!)^{b-1} \sum_{k=0}^{b-1} \alpha_{1} k p-((p-1)!)^{b} \\
\equiv & 0\left(\bmod p^{3}\right) .
\end{aligned}$$

Therefore, the maximum integer $c=3$.