Results 1 to 8 of 8

Math Help - Permutations.

  1. #1
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Permutations.

    Hello

    This is my first post.

    The question I'm working on is, 'All the digits of a number are different, the first digit is not zero, and the sum of the digits is 36. There are N x 7! such numbers. What is the value of N?'

    I think I have figured out that 1,9,2,8,3,7,6 gives me 7! Then there is 3,7,4,6,2,8,1,5, which gives me 8!. I can add and factor to get 9 x 7!. Now I need to include the zero. That gives me both 8 and 9 digit numbers but the zero can't be in the first place. I haven't studied permutations and am not sure how to proceed. I think I would do 8! and 9! respectively and subtract something but I can't get my head around that difference.

    If someone could point me in the right direction it would be very much apprecaited.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,639
    Thanks
    1592
    Awards
    1

    Re: Permutations.

    Quote Originally Posted by Furyan View Post
    'All the digits of a number are different, the first digit is not zero, and the sum of the digits is 36.
    I need some clarification.
    The digits 9876510 add up to 36.
    There are 6\cdot 6! arrangements of those digits that do not start with 0.
    Is that what you are working on?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Permutations.

    Plato, thank you very much for taking the time to reply. It's very helpful. Yes that is what I am working on. I thought that perhaps I had not yet found all the combinations of different digits with a sum of 36, but the zero was bothering me more, thank you for the pointer. I will check I haven't missed any other combinations. May I ask, is the number of arrangements of n different digits, when one of the digits cannot be in one place, (n-1) (n-1)!?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,639
    Thanks
    1592
    Awards
    1

    Re: Permutations.

    Quote Originally Posted by Furyan View Post
    May I ask, is the number of arrangements of n different digits, when one of the digits cannot be in one place, (n-1) (n-1)!?
    The longest string is 876543210 and there are 8\cdot8!~ rearrangements which do not have a leading zero.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Permutations.

    Plato, thank you for very much for all your help. Am I right in thinking that 9876510, from your previous post, is the shortest string?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,639
    Thanks
    1592
    Awards
    1

    Re: Permutations.

    Quote Originally Posted by Furyan View Post
    Plato, thank you for very much for all your help. Am I right in thinking that 9876510, from your previous post, is the shortest string?
    No you are not.
    The string 987651 also works.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Permutations.

    I see. I am extremely grateful your help. I will check with you once I think I have found the value of N.

    Thank you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Permutations.

    Hello Plato

    Thanks to you I was able to finish the problem.
    I certainly had missed a few combinations and should have given it more thought before posting originally!

    I got:

    87654321
    9876321
    9875421
    9865431
    9765432
    987651
    987642
    987543

    8! + (4)(7!) + (3)(6!)

    876543210
    98763210
    98754210
    98654310
    97654320
    9876510
    9876420
    9875430

    (8)8! + (28)(7!) + (18)(6!)

    Arrangements:

    (9)8! + (32)(7!) + (21)(6!)

    7![(9)(8) +32 +3]

    7!(107)

    N = 107

    Thanks again for your help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutations
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 12th 2010, 06:48 AM
  2. Permutations
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 9th 2010, 10:24 PM
  3. Permutations Help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 9th 2010, 07:30 PM
  4. permutations
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: February 11th 2009, 05:42 PM
  5. Permutations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 16th 2008, 06:59 PM

Search Tags


/mathhelpforum @mathhelpforum