Combinatroics 4-permuations of positive integers

[farleyknight]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

I've already taken a look at the book's answers, but I don't seem how they arrived at them.

First off there can only be 98 possible values for k. Next, there is 5 locations for the alternate integer to be, and there are 96 integers left to choose from. So for part a) my guess was 98*96*5=47040. However, the book gives the answer: 37927, which isn't even divisible by 5..

For part b), I got a similar answer. 2 positions this time: 98*96*2 = 18816, but this time I was closer to the books: 18,915, which isn't divisible by 2.

My only guess is that I'm misreading the problem.. But I'm not sure what's the correct interpretation.

 

[Avodyne]

Next, there is 5 locations for the alternate integer to be, and there are 96 integers left to choose from.

This should be 4 and 97, which gets you closer to the book's answer.

But some permutations have been counted more than once. Can you see why?