Results 1 to 2 of 2

Math Help - Permutations

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    Permutations

    How many 7-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6?

    I understand this part:

    There are four cases:

    1 is the first digit, 6 is the last digit << Case 1

    2 is the first digit, 6 is the last digit<< Case 2

    2 is the first digit, 2 is the last digit<< Case 3

    1 is the first digit, 2 is the last digit <<Case 4

    But. This is the part I don't understand...
    Why is it that for case 1, the answer is
    5! / (2!2!) ?

    How do we get the 5 and why do we use 5 and not 7? I understand the denominator, since there are duplicates of 2 AND 5, but why 5 in the numerator and not 7?

    For case 2 to 4, the answer is

    5! / 2!

    Why is it 5 and not 7 again? and how come this time there is only one duplicate taken into consideration in the denominator? What happens to the other one?

    Please explain this as clearly as possible. Thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,690
    Thanks
    617
    Hello, skeske1234!

    How many 7-digit even numbers less than 3,000,000 can be formed
    using the following digits: 1, 2, 2, 3, 5, 5, 6 ?

    I understand this part:

    There are four cases:

    1 is the first digit, 6 is the last digit << Case 1

    2 is the first digit, 6 is the last digit << Case 2

    2 is the first digit, 2 is the last digit << Case 3

    1 is the first digit, 2 is the last digit << Case 4


    But this is the part I don't understand . . .
    Why is it that for case 1, the answer is: \frac{5!}{2!\,2!} ?

    How do we get the 5 and why do we use 5 and not 7?
    I understand the denominator, since there are duplicates of 2 and 5,
    but why 5 in the numerator and not 7?

    In Case 1, the first digit is 1, the last is 6: .1 _ _ _ _ _ 6

    And we must place the other five digits {2, 2, 3, 5, 5} in those spaces.
    . . And there are: . \frac{5!}{2!\,2!} ways.


    In Case 3, the first digit is 2, the last is 2: .2 _ _ _ _ _ 2

    And we place the other five digits {1, 3, 5, 5, 6} in those spaces.
    . . And there are: . \frac{5!}{2!} ways.


    Get the idea?

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutations
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 13th 2009, 05:10 AM
  2. Permutations
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: November 26th 2009, 09:55 PM
  3. Permutations
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 28th 2009, 02:27 PM
  4. permutations
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: May 18th 2009, 01:38 AM
  5. Permutations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 16th 2008, 06:59 PM

Search Tags


/mathhelpforum @mathhelpforum