How many 4-permutations of the positive integers not exceeding $100$ contain three consecutive integers in the correct order
a.) where consecutive means in the usual order of the integers and where these consecutive integers can perhaps be separated by other integers in the permutation?
b.) where consecutive means both that the numbers be consecutive integers and that they be in consecutive integers and that they be in consecutive positions in the permutation?
3 consecutive integers is, $k, k+1, k+2$, thus $k$ is limited to $98$,
a.) The number of ways we can place and the 2 others is $C(4, 3) = 4$. Since $3$ integers are already taken by $k, k+1, k+2$, the last integer can have 97 integers, thus
$$4*98*97 = 38,024$$
b.) Since $k, k+1, k+2$ have to be consecutive in placement, there's only two possible place for that, thus
$$2*98*97 = 19, 012$$
Both of my answers are a bit off compared to the answer key. In the answer key, a.) $37,927$ b.) $18,915$. One might notice that if my answers are subtracted $97$, I get the answer key's answer. So how exactly do you arrive to these numbers? I'm out of ideas.