The first question is relatively simple, and the following is 2 3 4 1 or 2 3 1 4.

The second question, f 10 can only take 1 or 10, but when f 1= 1, it can be deduced that fm=m, so f 1 is not equal to 1, and the minute.

C8=56, and the other three unequal numbers are b, c, d C and d, so their arrangement can only be 1bc D.

B c d 1, so there are 56 situations.

When f 10 is equal to 1, f 1 is not equal to 1, so 6 out of 2 to 9 are equal to themselves, and the total number is 6.

C8=28, the other two unequal numbers are b and c, and their arrangement can only be1b c.

B c 10, so there are 28 cases, plus 56+28=84, so the number of optimal mapping sets is 84.