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Permutations (easy) ?

  1. May 16, 2012 #1
    Just need to learn how to use permutations in this question

    How many integers between 100 and 150 have three different digits in increasing order.
    One such is 147.

    I solved this using a long and unnecessary method.

    Can anyone guide me in how to solve it using permutations.
    I got 18 as my answer.
     
  2. jcsd
  3. May 16, 2012 #2

    coolul007

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    Gold Member

    I get 32 numbers, the tens digit can contain 4 values, 0,2,3,4. While the ones digit can contain 8 values.
     
  4. May 16, 2012 #3
    Well I don't see how you can plug this into a permutation equation and get an answer, but you can easily work it out analytically.

    Rules:
    1. The digits must be in increasing order.
    2. The second and third digit must be greater than 1.
    3. The second digit must be greater than 1 but less than 5.
    4. The third digit must be greater than the second.

    So that only leaves us with 2, 3, and 4 for the second digit.

    When the second digit is 2 you can have 9-2 = 7 numbers in that have digits that are increasing: 23, 24, 25, 26, 27, 28, 29. When the second digit is 3 you can have 9-3 = 6 numbers: 34, 35, 36, 37, 38, 39. When the second digit is 4 you can have 9 - 4 = 5 numbers: 45, 46, 47, 48, 49. So in total you have 7 + 6 + 5 = 18 numbers. You are right.
     
  5. May 19, 2012 #4
    12a or 13b or 14c
    a>2 , b>3 , c>4
    7 ways or 6 ways or 5 ways
    18 ways.
     
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