7・122 For each permutation $\tau=\left(x_{1}, x_{2}, \cdots, x_{10}\right)$ of $1,2, \cdots, 10$, define
$$
S(\tau)=\sum_{k=1}^{10}\left|2 x_{k}-3 x_{k+1}\right| \text {, }
$$

and assume $x_{11}=x_{1}$. Try to find
(1) the maximum and minimum values of $S(\tau)$;
(2) the number of all permutations $\tau$ that make $S(\tau)$ reach its maximum value;
(3) the number of all permutations $\tau$ that make $S(\tau)$ reach its minimum value.
[Solution] (1) Write down the 20 numbers which are the 2 times and 3 times of $1,2, \cdots, 10$ as follows:
$20,18,16,14,12,10,8,6,4,2$,
$$
30,27,24,21,18,15,12,9,6,3
$$

The difference between the sum of the 10 larger numbers and the sum of the 10 smaller numbers is $203-72=131$. Therefore, $S(\tau) \leqslant 131$. For the permutation
$$
\tau_{0}=(1,5,6,7,2,8,3,9,4,10),
$$

it is easy to calculate that $S\left(\tau_{0}\right)=131$, so the maximum value of $S(\tau)$ is 131.
To estimate the minimum value of $S(\tau)$, we should select the largest 10 numbers from the 20 numbers in (2) as subtrahends. Clearly, 30, 27, 24, and 21 cannot be selected, while 3 and 2 must be selected. The largest numbers that can be selected are 20 and 18, and only one of the two 18s can be selected. Subsequently, the largest number that can be selected is 16, which means 15 cannot be selected. Proceeding in this manner, the largest 10 subtrahends are 20, 18, 16, 14, 12, 10, 8, 6, 3, 2. Therefore, $S(\tau) \geqslant 57$. For the permutation
$$
\tau_{1}=(10,9,8,7,6,5,4,3,2,1),
$$

it is easy to calculate that $S\left(\tau_{1}\right)=57$. So the minimum value of $S(\tau)$ is 57.
(2) Among $2 x_{k}$ and $3 x_{k+1}$, the smaller one is called the small number, and the other is called the large number. Since the 8 numbers generated by 1, 2, 3, 4 must all be small numbers and the 8 numbers generated by 10, 9, 8, 7 must all be large numbers, in the permutation that maximizes $S(\tau)$, 1, 2, 3, 4 are not adjacent to each other, and 7, 8, 9, 10 are also not adjacent to each other. 5 and 6 cannot immediately follow 7, 8, 9, 10, nor can they immediately precede 1, 2, 3, 4.
Assume $x_{1}=1$, and refer to the following symbol arrangement
$$
1 \triangle \bigcirc \square \triangle \bigcirc \square \triangle \bigcirc \square \triangle \bigcirc \text {. }
$$

2, 3, 4 can be arbitrarily filled into 3 $\square$s, with 6 different ways; 7, 8, 9, 10 can be arbitrarily filled into 4 circles, with 24 different ways; 5 can be filled into one of the 4 $\triangle$s, with 4 different ways; 6 can be filled into one of the 4 $\triangle$s, and when it is in the same $\triangle$ as 5, it can be either before or after 5, with 5 different ways. In summary, when $x_{1}=1$, the number of different permutations that maximize $S(\tau)$ is
$$
6 \times 24 \times 4 \times 5=2880 .
$$

Thus, by cyclic permutation, the total number of different permutations that maximize $S(\tau)$ is 28800.
(3) In the discussion of (1), it is known that to minimize $S(\tau)$, the 10 large numbers should be 30, 27, 24, 21, 18, 15, 12, 9, 6, 4. To make 4 a large number, 2 should be placed before 1; to make 6 a large number, 3 should be placed before 2; to make 9 a large number, 4 should be placed before 3; to make 12 a large number, 4 should be preceded by 5 or 6, so in the minimum permutation, there must be 5 consecutive terms as $5,4,3,2,1$ or $6,4,3,2,1$. The following refers to the permutation that minimizes $S(\tau)$ as the minimum permutation.

Assume $x_{1}=10$ and consider the minimum permutation containing 5 consecutive terms as $5,4,3,2,1$. To make 15 and 18 large numbers, 5 and 6 cannot immediately follow 10, i.e., $x_{2}$ can only be one of 7, 8, 9. 6, 7, 8, 9, these 4 numbers, one of them is $x_{2}$, and the other 3 numbers can be arranged arbitrarily after $x_{2}$, with 6 ways for each case, totaling 18 permutations. At this time, the 5 consecutive terms $5,4,3,2,1$ can be placed after 6 or 7. Thus, a total of 36 different permutations are obtained.

Next, consider the minimum permutation containing 5 consecutive terms as $6,4,3,2,1$. At this time, the remaining 4 numbers are 5, 7, 8, 9. They have 24 different permutations, of which 6 permutations starting with 5 do not meet the requirements; among the 6 permutations where 5 is after 9 and the 6 permutations where 5 is after 8, only the 6,4,3,2,1 can be placed before 5, and 9 and 8 cannot be separated by 5. A total of 12 permutations are obtained. The remaining 6 permutations are 5 adjacent to 7. At this time, 6,4,3,2,1 can be placed after any of the 4 numbers, totaling 24 permutations. Therefore, the minimum permutations containing 6,4,3,2,1 total 36.

In summary, the minimum permutations with $x_{1}=10$ total 72. By cyclic permutation, the total number of minimum permutations that minimize $S(\tau)$ is 720.