$1 \cdot 411,2,3,4,5,6$ 的排列 $\left(n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}\right)$ 具有如下性质: 对于 $i=1,2,3,4,5,\left(n_{1}, n_{2}, \cdots, n_{i}\right)$ 都不是 $(1,2, \cdots, i)$ 的排列. 求这种排列的个数.

The permutation $\left(n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}\right)$ of $1, 2, 3, 4, 5, 6$ has the following property: for $i=1,2,3,4,5, \left(n_{1}, n_{2}, \cdots, n_{i}\right)$ is not a permutation of $(1,2, \cdots, i)$. Find the number of such permutations.
[Solution 1] Clearly, $n_{1} \neq 1$.
For $n_{1}=6$, all $5!$ permutations satisfy the requirements of the problem.
For $n_{1}=5$, out of the $5!$ permutations, $4!$ permutations where $n_{6}=6$ do not satisfy the requirements, so the number of permutations that satisfy the requirements is $5!-4!$.
For $n_{1}=4$, out of the $5!$ permutations, those where $n_{6}=6$ or $n_{6}=5, n_{5}=6$ do not satisfy the requirements, so the number of permutations that satisfy the requirements is $5!-4!-3!$.
For $n_{1}=3$, out of the $5!$ permutations, those of the form $3 \times \times \times \times 6, 3 \times \times \times 65, 3 \times \times 645, 3 \times \times 564, 3 \times \times 654$ do not satisfy the requirements, so the number of permutations that satisfy the requirements is $5! - 4! - 3! - 3 \times 2!$.
For $n_{1}=2$, out of the $5!$ permutations, those of the form $2 \times \times \times \times 6, 2 \times \times \times 65, 2 \times \times 645, 2 \times \times 564, 2 \times \times 654, 21 \times \times \times 3, 21 \times \times 34, 21 \times \times 35, 21 \times 3 \times 4, 216345$ do not satisfy the requirements, so the number of permutations that satisfy the requirements is $5! - 4! - 2 \times 3! - 6 \times 2! - 1$.
In summary, the number of permutations that satisfy the requirements is
$$
\begin{array}{l}
5! + (5! - 4!) + (5! - 4! - 3!) + (5! - 4! - 3! - 3 \times 2!) + (5! \\
- 4! - 2 \times 3! - 6 \times 2! - 1) \\
= 461 \text{.}
\end{array}
$$
[Solution 2] We will calculate the number of permutations that do not satisfy the requirements of the problem. Let $\left(m_{1}, m_{2}, m_{3}, m_{4}, m_{5}, m_{6}\right)$ be a permutation that does not satisfy the requirements. Thus, there exists $2 \leqslant i \leqslant 6$ such that
$$
\begin{array}{l}
\left\{m_{i}, m_{i+1}, \cdots, m_{6}\right\} = \{i, i+1, \cdots, 6\}, \\
\left\{m_{j}, m_{j+1}, \cdots, m_{6}\right\} \neq \{j, j+1, \cdots, 6\}, \quad j = i+1, \cdots, 6,
\end{array}
$$

where when $i=6$, (2) does not exist. We denote the number of permutations $\left(m_{i}, m_{i+1}, \cdots, m_{6}\right)$ that satisfy (1) and (2) by $f(i)$, then we have
$$
\begin{array}{l}
f(6) = 1, \quad f(5) = 2! - f(6) = 1, \\
f(4) = 3! - f(5) - 2! f(6) = 3, \\
f(3) = 4! - f(4) - 2! f(5) - 3! f(6) = 13, \\
f(2) = 5! - f(3) - 2! f(4) - 3! f(5) - 4! f(6) = 71.
\end{array}
$$

Thus, the number of permutations that do not satisfy the requirements is
$$
S = \sum_{i=2}^{6} f(i) \cdot (i-1)! = 259.
$$

Therefore, the number of permutations that satisfy the requirements is $6! - 259 = 461$.