Results 1 to 6 of 6

Thread: a wee puzzle

  1. #1
    Newbie
    Joined
    Feb 2019
    From
    paisley
    Posts
    3

    a wee puzzle

    A club receives 92 applications for 49 available spaces, the applications are received from 3 different groups, group 1 consists of 68 applicants, group 2 consists of 16 applicants, group 3 consists of 8 applicants.
    All of the applicants are allocated a number 1-92, placed into a hat & drawn out randomly
    What are the odds of 1/2 of each group being successful applicants?
    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: a wee puzzle

    With 92 applicants, divided into groups of 68, 16, and 8, "1/2 of each group" would be 34+ 8+ 4= 46, not 49, so that is impossible. Did you mean "at least half"?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2019
    From
    paisley
    Posts
    3

    Re: a wee puzzle

    Apologies, yes, at least 1/2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,362
    Thanks
    3253
    Awards
    1

    Re: a wee puzzle

    Quote Originally Posted by oldno7 View Post
    A club receives 92 applications for 49 available spaces, the applications are received from 3 different groups, group 1 consists of 68 applicants, group 2 consists of 16 applicants, group 3 consists of 8 applicants.
    All of the applicants are allocated a number 1-92, placed into a hat & drawn out randomly
    What are the odds of 1/2 of each group being successful applicants?
    Thanks in advance
    As corrected to read "at least one-half of each group".
    This calculation is for one half plus one of each group. SEE HERE.
    I chose to do that one because it nicely assigns all available memberships.
    But you can see there are many possible assignments for "at least" half.
    Assigning exactly half lives three memberships unassigned. Those three can be assigned to three groups in ten ways.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: a wee puzzle

    Since 49- 46= 3 this is just a matter of determining how many ways you can allocate 3 slots to 3 different groups. There are 3 ways to give all three slots to one group. There are 6 ways to allocate 2 of the slots to one group and one to the other. There are 3!= 6 ways to one slot to each group.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,362
    Thanks
    3253
    Awards
    1

    Re: a wee puzzle

    Quote Originally Posted by HallsofIvy View Post
    Since 49- 46= 3 this is just a matter of determining how many ways you can allocate 3 slots to 3 different groups. There are 3 ways to give all three slots to one group. There are 6 ways to allocate 2 of the slots to one group and one to the other. There are 3!= 6 ways to one slot to each group.
    If we count the number of ways to arrange the string of two bars and three stars, $||***$, that gives the number of ways to place three memberships into three groups.
    That is the arrangement $*|~|**$ would mean group one gets one, group two gets none, and group three gets two.
    As I said above it is $\dfrac{5!}{3!\cdot 2!}=10$ ways.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with puzzle
    Posted in the Math Puzzles Forum
    Replies: 0
    Last Post: Jan 6th 2012, 08:26 AM
  2. Help with a puzzle
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Dec 13th 2011, 02:00 PM
  3. puzzle
    Posted in the Geometry Forum
    Replies: 5
    Last Post: Aug 18th 2011, 05:20 AM
  4. Need Help With This Puzzle.... Thanks!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 8th 2009, 09:35 AM
  5. A puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: Oct 29th 2007, 11:59 PM

/mathhelpforum @mathhelpforum