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Math Help - easy combination/probability problem

  1. #1
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    easy combination/probability problem

    my problem is:
    How many 5 digit pin number combinations can you have if there are no repeated digits with numbers 0-9.
    The first answer i had i was confident in but it was wrong i believe. I had:
    10*9*8*7*6*5/5*4*3*2*1
    this is because a pin # is a 5 element subset of 10 digits. But, this is probably wrong because i was forgetting that the numbers can be in any order, not in just any combination.
    This would be correct if the problem was there are 10 people and you must pick 5 names to write down and they didnt have to be in any specific order.


    I added the last example for clarity.
    thanks.

    EDIT:
    I think i might have it solved, but can someone make sure it is right?
    10!/5!.
    But i have one more problem but i might also have it solved can someone check it?
    if there must be the number 2 in each pin #, with the same constraints as the problem above.
    i beleive the answer would be 10!/4! or 9!/4! . this would of course be wrong if the first problem is wrong.

    thanks.
    Last edited by ihaveamathproblem; November 22nd 2007 at 07:58 PM. Reason: another problem
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ihaveamathproblem View Post
    my problem is:
    How many 5 digit pin number combinations can you have if there are no repeated digits with numbers 0-9.
    The first answer i had i was confident in but it was wrong i believe. I had:
    10*9*8*7*6*5/5*4*3*2*1
    this is because a pin # is a 5 element subset of 10 digits. But, this is probably wrong because i was forgetting that the numbers can be in any order, not in just any combination.
    This would be correct if the problem was there are 10 people and you must pick 5 names to write down and they didnt have to be in any specific order.


    I added the last example for clarity.
    thanks.

    EDIT:
    I think i might have it solved, but can someone make sure it is right?
    10!/5!.
    But i have one more problem but i might also have it solved can someone check it?
    if there must be the number 2 in each pin #, with the same constraints as the problem above.
    i beleive the answer would be 10!/4! or 9!/4! . this would of course be wrong if the first problem is wrong.

    thanks.
    the first can be any of 10
    the second any of 9
    the third any of 8
    the fourth any of 7
    the fith any of 6

    So total is 10x9x8x7x6 = 10!/5!

    RonL
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