Discrete Math and its applications by Kenneth Rosen 7th edition Section 6:Counting

Question

Can this question be explained in detail as i have trouble understanding it? "How many 4-permutations of the positive integers not exceeding 100 contain three consecutive integers k, k + 1, k + 2, in the correct order" a) where these consecutive integers can perhaps be separated by other integers in the permutation? b) where they are in consecutive positions in the permutation?

for part a the solution is 98*97*4-97, the solution manual says we have to account for double counting but I don't understand how? Why 97?

Answer 1

There are 98*97*4 ways to choose three consecutive integers, one ‘other’ number, and finally, permute them. But there is an overcounting of the permutations where you choose 4 consecutive integers. Suppose, out of 98 ways to select three consecutive integers, you select, 1,2,3. And then for the other integer you select 4. Now consider the permutation 1,2,3,4. This same permutation will be counted again when you select three consecutive integers to be 2,3,4 and one other number to be 1. Therefore, to account for the overcounting of 4 consecutive integers you need to subtract 97 to get the correct answer.