2. $\frac{1}{2} n(n+1)$ distinct numbers are randomly arranged in a triangle:

Let $M_{\mathrm{k}}$ be the maximum number in the $k$-th row (counting from the top), find the probability that $M_{1}<M_{2}<M_{3}<\cdots<M_{\mathrm{n}}$ holds.
2. Let the required probability be $p_{\mathrm{n}}$, obviously, $p_{1}=1$, $p_{2}=2 / 3$.

Since the largest number must appear in the last row to satisfy the inequality described in the problem. The probability of this situation occurring is
$$
\frac{n}{\frac{1}{2} n(n+1)}=\frac{2}{n+1} .
$$

Therefore,
$$
p_{n}=\frac{2}{n+1} p_{n-1}=\cdots=\frac{2^{n}}{(n+1)!} \quad(n \geqslant 2) .
$$