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Math Help - Permutations and combinations Help

  1. #1
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    Permutations and combinations Help

    Sorry guys one more question i need help on. Thanks a lot. I really appreciate this. The question is:

    There are 6 numbers, drawn at random, between 1-49. Once a number is chosen, it no longer can be used again. Each ticket has 6 numbers.

    Grand prize:all numbers match the 6 numbers drawn
    2nd prize: 5 numbers match
    3rd prize: 4 numbers match
    4th prize: 3 numbers match

    How many ways are there for selecting the numbers on a ticket that can win the grand prize?
    I do the same to find out for the 2nd, 3rd, and 4th prizes.

    Thank you for any help given.
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  2. #2
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    Hi.

    Quote Originally Posted by ajmaal14 View Post
    Sorry guys one more question
    You don't need to feel sorry for that

    Quote Originally Posted by ajmaal14 View Post
    The question is:

    There are 6 numbers, drawn at random, between 1-49. Once a number is chosen, it no longer can be used again. Each ticket has 6 numbers.

    Grand prize:all numbers match the 6 numbers drawn
    2nd prize: 5 numbers match
    3rd prize: 4 numbers match
    4th prize: 3 numbers match

    How many ways are there for selecting the numbers on a ticket that can win the grand prize?
    I do the same to find out for the 2nd, 3rd, and 4th prizes.
    Well, at first you draw one of the numbers - there are 49 different balls;
    you don't use this number again, so there are (49-1)=48 balls left
    2. draw: you draw one number out of 48 => 48 possibilities for that
    You do not use that number again, so there are 47 numbers left
    3. draw: you draw one out of 47 (47 possibilities)
    4. draw: one out of 46
    5. draw: one out of 45
    6. draw: one out of 44

    So there are 49*48*47*46*45*44 different ways of drawing 6 numbers out of 49.

    Edit: I made a mistake. Sorry
    Plato gave the right answer: Because of
    A lottery ticket is treated as a combination not a permutation, order makes no difference.
    you need to divide by 6!, thus
    \frac{49*48*47*46*45*44}{6*5*4*3*2*1} = \begin{pmatrix} 49 \\ 6 \end{pmatrix}
    is the solution!


    Yours
    Rapha
    Last edited by Rapha; June 19th 2009 at 10:46 AM. Reason: Restored original reply - Mistakes, when corrected, can still be useful and serve a teaching purpose.
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  3. #3
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    Quote Originally Posted by Rapha View Post
    By the way, it is
    49*48*47*46*45*44 = \begin{pmatrix} 49 \\ 6 \end{pmatrix}
    Actually it should be \frac{49*48*47*46*45*44}{6*5*4*3*2*1} = \begin{pmatrix} 49 \\ 6 \end{pmatrix}
    A lottery ticket is treated as a combination not a permutation, order makes no difference.
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