II. (Problem 16) The permutation $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ of $1,2,3,4,5$ has the following property: for $1 \leqslant i \leqslant 4$, $a_{1}$, $a_{2} \cdots, a_{i}$ does not form a permutation of $1,2, \cdots, i$. Find the number of such permutations.
Obviously, $a_{1} \neq 1$.
When $a_{1}=5$, all $4!$ permutations meet the requirements.
When $a_{1}=4$, the $3!$ permutations where $a_{5}=5$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!)$.

When $a_{1}=3$, permutations in the form of $3 \times \times \times 5$ and $3 \times \times 54$ do not meet the requirements, so the number of permutations that meet the requirements is $(4!-3!-2!)$.

When $a_{1}=2$, permutations in the form of $2 \times \times \times 5, 2 \times \times 54, 2 \times 534, 2 \times 453, 2 \times 543$ do not meet the requirements, (continued on page 43)
(continued from page 36) so the number of permutations that meet the requirements is $(4! - 3! - 2! - 3 \times 1! - 1! - 1!)$.
In summary, the number of permutations that meet the requirements is
$$
\begin{array}{l}
4!+(4!-3!)+(4!-3!-2!)+(4!-3!-2! \\
-3 \times 1!)=71
\end{array}
$$