Results 1 to 2 of 2

Math Help - Combination Problem... Help Plz

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    3

    Combination Problem... Help Plz

    Question Details:
    Consider the set of 6-digit integers, where leading 0's are permitted. Two integers are considered to be "equivalent" if one can be obtained from the other by a rearrangement (permutation) of the digits. Thus 129450 and 051294 are "equivalent". Among all the 10^6 six-digit integers:

    a. how many non-equivalent integers are there?
    b. if digits 0 and 9 can appear at most once, how many non-equivalent integers are there?
    c. Generalize your results to (a) and (b) for n-digit integers.

    My brain is going to explode~~ Thank you for the helps.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, jmmjm!

    This seemed like a formidable problem,
    . . but with a little thinking . . .



    Consider the set of 6-digit integers, where leading 0's are permitted.
    Two integers are considered to be "equivalent" if one can be obtained
    from the other by a rearrangement (permutation) of the digits.
    Thus 129450 and 051294 are "equivalent".

    Among all the 10^6 six-digit integers:

    a) how many non-equivalent integers are there?
    Nearly all six-digit numbers have an equivalent "mate" somewhere in the list.

    Example: .000007 has permutations: 000070, 000700, 007000, 070000, 700000


    The only six-digit numbers which do not have equivalent mates
    . . are those consisting of exactly one digit.

    Hence: .000000, 111111, 222222, ..., 999999 are non-equivalent.

    Therefore, there are ten non-equivalent integers.



    b) if digits 0 and 9 can appear at most once, how many non-equivalent integers are there?
    Since 0 and 9 cannot appear six times,

    . . only: .111111,222222, ..., 888888 are non-equivalent.

    Therefore, there are eight non-equivalent integers.



    c) Generalize your results to (a) and (b) for n-digit integers.
    (a) For n-digit numbers, the non-equivalent integers are:
    . . . . 000...0, .111...1, .222...2, .. . . , .999...9

    . . The answer is still ten.


    (b) Similarly, the answer is still eight.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. combination problem need help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 9th 2011, 03:47 PM
  2. Combination Problem
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 19th 2008, 03:54 AM
  3. Combination Problem
    Posted in the Statistics Forum
    Replies: 4
    Last Post: October 19th 2008, 08:31 AM
  4. Combination problem
    Posted in the Statistics Forum
    Replies: 4
    Last Post: May 18th 2008, 01:59 PM
  5. Combination problem??
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 14th 2008, 09:25 PM

Search Tags


/mathhelpforum @mathhelpforum