1.48 Arrange the natural numbers $3,4,5, \cdots, 1994,1995$ into a sequence $\left\{a_{k}\right\}$, such that
$$
k \mid a_{k}, \quad k=1,2, \cdots, 1993 .
$$

How many different arrangements are there?
［Solution］Except for 1994 and 1995, every natural number can be placed in a position with its own number. Therefore, the first step is to arrange the positions of 1994 and 1995.

From (1), we know that 1994 and 1995 can only be placed in positions with their own divisors as numbers. Thus, this divisor cannot be placed in a position with its own number, and it must be placed in a position with one of its divisors as a number (which is also a divisor of the original number). This process continues until the last divisor is placed in position 1 or 2.
Since
$$
1995=3 \times 5 \times 7 \times 19, \quad 1994=2 \times 997,
$$

the substitution process starting from 1995 must end with the last divisor being placed in position 1. Therefore, 1994 can only be placed in position 2. This means that each arrangement that meets the requirements corresponds to a permutation of some divisors of 1995, and this correspondence is a bijection, i.e., a one-to-one mapping. In fact, other numbers can only be placed in positions with their own numbers. Otherwise, it would lead to a smaller number being placed in a larger position, which of course does not meet the requirement of (1).
Arrange the divisors from largest to smallest, and only record the multiples of adjacent divisors. For example, the sequence 1995 $\longrightarrow 665 \longrightarrow 35 \longrightarrow 7 \longrightarrow 1$ is recorded as $(3,19,5,7)$. Clearly, the product of the terms in each permutation is 1995.
Consider all permutations starting with 3, there are 13 such permutations:
$$
\begin{array}{l}
(3,5,7,19),(3,5,19,7),(3,5,7 \times 19), \\
(3,7,5,19),(3,7,19,5),(3,7,5 \times 19), \\
(3,19,5,7),(3,19,7,5),(3,19,5 \times 7), \\
(3,5 \times 7,19),(3,5 \times 19,7),(3,7 \times 19,5), \\
(3,5 \times 7 \times 19) .
\end{array}
$$

Similarly, there are 13 permutations for each of the starting terms 5, 7, and 19. Therefore, the number of such permutations (with the first term being a prime factor) is $13 \times 4 = 52$.
In addition, there are 3 permutations for the first term being $3 \times 5$:
$$
(15,7,19),(15,19,7),(15,7 \times 19).
$$

Similarly, there are 3 permutations for each of the starting terms $3 \times 7, 3 \times 19, 5 \times 7, 5 \times 19, 7 \times 19$, totaling $6 \times 3 = 18$.
Finally, there are 5 permutations:
$$
(3 \times 5 \times 7,19),(3 \times 5 \times 19,7),(3 \times 7 \times 19,5),
$$
$(5 \times 7 \times 19,3),(1995)$.
In summary, the total number of permutations that meet the requirements is $52 + 18 + 5 = 75$.