Results 1 to 3 of 3

Math Help - thee permutations and combinations question....please help

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    2

    thee permutations and combinations question....please help

    1)how many numbers between 2500-6500
    using 1,2,3,4,5,6,7 without any repitition

    i answered
    5P1 3P1 7P1 7P1=735

    2)Jack needs to answer 5 out of 8 questions find the number of combinations of answers if he can answer any 5 questions

    i answered
    8C5=56

    b)if he has to answer 3 out of 4 in paper A and 2 out of 4 in paper B

    i answered
    4C3 4C2= 24

    3)6 members must be chosen from 7 boys and 5 girls
    a) if there are no restrictions
    i answered
    12C6=924

    b)there are not more than 2 girls
    5C2 7C4= 350

    c) there are 4 girls
    5C4 7C2=105

    could anyone verify if i am right or wrong.....and please tell me where i went wrong....please
    Last edited by cantcount; June 3rd 2010 at 02:06 PM. Reason: i forgot to mention i am unsure of the answers, mind verifying for me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,666
    Thanks
    1617
    Awards
    1
    Quote Originally Posted by cantcount View Post
    1)how many numbers between 2500-6500
    using 1,2,3,4,5,6,7 without any repitition
    i answered: 5P1 3P1 7P1 7P1=735
    This problem is more complicated than you have considered.
    If the number should begin with 2 then the second digit must be 5, 6, or 7.
    The last two digits can be _5P_2=20. That gives 60 numbers begin with 2.
    If the first digit is 6, then the second digit can be 2, 3, or 4. SO?
    If the first digit is 3, 4, or 5, then the next three can be _6P_3=120. SO?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Hello, cantcount!

    3) 6 members must be chosen from 7 boys and 5 girls

    . . b) There are not more than 2 girls

    "Not more than 2 girls" means: .0 girls, 1 girl, or 2 girls.


    . . \begin{array}{ccccccc}\text{0 girls, 6 boys:} & (_5C_0)(_7C_6) &=& 1\cdot 7 &=& \;\;7 \\ \text{1 girl, }\text{ 5 boys:} & (_5C_1)(_7C_5) &=&5\cdot21 &=& 105 \\ \text{2 girls, 4 boys:} & (_5C_2)(_7C_4) &=& 10\cdot35 &=& 350 \\ \hline & & & \text{Total:} && 462 \end{array}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutations of Combinations question? Help
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: September 15th 2010, 09:46 AM
  2. Permutations/Combinations question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 7th 2010, 10:32 AM
  3. Permutations / Combinations Question
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: July 19th 2009, 09:42 PM
  4. Combinations/ Permutations question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 30th 2009, 01:31 AM
  5. another permutations / combinations question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 26th 2008, 06:02 PM

Search Tags


/mathhelpforum @mathhelpforum