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Permutation Question.

  1. Dec 15, 2005 #1
    The question is:

    A sequence of length 6 is formed from the digits {0,1,2...9}. If no repetition is allowed, how many of these sequences can be formed if:

    f) the sum of the first two terms is 7?

    So i set up my place holders:

    _ _ _ _ _ _

    if the first 2 place holders have a sum of 7, are there 8 possibilities for each place holder because 0+1, 1+6, 2+5, 3+4, 4+3, 5+2, 6+1, 7+0? Or would there be 4 for each consisting only of 0+1, 1+6, 2+5, 3+4? So really teh answer would be 16 x 8P4 or 64 x 8P4?

    The answer was 13440, but I don't see why.
     
  2. jcsd
  3. Dec 15, 2005 #2

    Dale

    Staff: Mentor

    You are right on the right track. There are 8 possibilities for the first digit (0-7). Once that is picked the second digit is determined. That leaves 8 possible digits for the third, 7 possible for the fourth, 6 possible for the fifth, and 5 possible for the sixth.
    8*1*8*7*6*5
    8*8P4
    13440

    -Dale
     
  4. Dec 17, 2005 #3
    I see, ok.

    Thanks!
     
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