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Combinatroics 4-permuations of positive integers

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    3. 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.
     
  2. jcsd
  3. Apr 3, 2009 #2

    Avodyne

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    Science Advisor

    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?
     
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