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Posers and Puzzles

Posers and Puzzles

  1. Standard member Surtism
    Prof.
    03 Oct '09 00:01
    .... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:

    "A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combination. I.E. How many nights will that one fairytale last?"

    As said, I got the answer, but used Excel to do it - please would somebody help me with a more technical way of attacking this problem.


    Many thanks.
    Surtism
  2. Subscriber joe shmo On Vacation
    Strange Egg
    03 Oct '09 01:15
    Originally posted by Surtism
    .... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:

    "A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combina ...[text shortened]... mebody help me with a more technical way of attacking this problem.


    Many thanks.
    Surtism
    I believe the formula you seek is

    C(n,r) = (n!)/(r!*(n-r)!)

    C(n,r) the number of combinations of "n" distinct objects taken "r" at a time.

    so your problem

    C(14,4) = 14!/(4!*(14-4)!)

    = (14*13*12*11*10!)/(4!*10!)

    =(14*13*12*11)/(4*3*2*1)

    =1001 days
  3. Standard member forkedknight
    Defend the Universe
    03 Oct '09 04:17
    Originally posted by Surtism
    .... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:

    "A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combina ...[text shortened]... mebody help me with a more technical way of attacking this problem.


    Many thanks.
    Surtism
    I don't quite follow your statement, "i.e. How many nights will that one fairytale last?"

    If the requirement is that each child can only hear the fairytale once, then there is simply 14/4 = 3.5 ~= 4 nights that the fairytale will last.

    If you did every combination, every child would hear the fairytale multiple times.
  4. 03 Oct '09 09:48
    ^ Interactive tale perhaps.
  5. Standard member Surtism
    Prof.
    05 Oct '09 22:00
    Originally posted by joe shmo
    I believe the formula you seek is

    C(n,r) = (n!)/(r!*(n-r)!)

    C(n,r) the number of combinations of "n" distinct objects taken "r" at a time.

    so your problem

    C(14,4) = 14!/(4!*(14-4)!)

    = (14*13*12*11*10!)/(4!*10!)

    =(14*13*12*11)/(4*3*2*1)

    =1001 days
    Thanks very much, so in simple terms:

    4/14 x 3/13 x 2/12 x 1/11 = 1/1001


    How would it work for the probability of getting 3 numbers on the UK lottery?
  6. Subscriber joe shmo On Vacation
    Strange Egg
    06 Oct '09 03:20
    Originally posted by Surtism
    Thanks very much, so in simple terms:

    4/14 x 3/13 x 2/12 x 1/11 = 1/1001


    How would it work for the probability of getting 3 numbers on the UK lottery?
    ahhh...i think you just replace "n" with how many numbers are on the ticket, and "r" would be "3" for your particular situation....

    I get

    49!/(3!*(49-3)!)=18424

    so 1/18424

    I could be wrong.