1. is first transformed into a problem of permutation and combination, and then all possible proportions in all combinations are calculated.

2. Suppose there are m authors and works respectively, the first author has m corresponding connection possibilities, and the second one is m- 1.

The middle possibility, the third possibility is m-3, until the last one is known, there is only 1 possibility.

3. The above possible products are 1 * 2 * 3 * 4 *...* m, and the permutation number of m. ..

5, there is only one correct result, and his probability is naturally easy to calculate.

6. Because it is impossible to make one mistake, there are at least two mistakes. In the case of two errors, we can think of it this way: two errors are randomly counted, and there are several results by randomly selecting two in M, that is, Cm2 (I'm sorry I can't express it here, but it is actually the combination of two numbers in M).

7. If three is wrong, according to the arrangement of three, P33 has one possibility of being right and C32 has two possibilities of being wrong, that is, three possibilities.

……

By analogy, the original intention of the topic is particularly clear and it is not difficult to answer!

I hope I can help you!