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Thread: different digits

  1. #1
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    different digits

    how many integers between 100 and 1000 have different digits.?

    im trying to solve this using permutations.
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  2. #2
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    Re: different digits

    We're talking a 3 digit number which obviously cannot start with the digit 0

    So we have 9 choices for the first digit, and then 9 for the second digit and 8 for the third

    for a total of 648 numbers.

    the fact that 0 cannot appear as the first digit is going to mess up your permutation calc.

    you could look at it as the numbers w/no 0 + the numbers with 1 0

    The first term is $3! \binom{9}{3}=504$

    The second term is $\binom{9}{1}2! \binom{8}{1} = 144$

    and these sum to $504+144=648$
    Last edited by romsek; Apr 13th 2017 at 07:16 AM.
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  3. #3
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    Re: different digits

    1: 102
    2: 103
    3: 104
    ....
    646: 985
    647: 986
    648: 987
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  4. #4
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    Re: different digits

    Quote Originally Posted by DenisB View Post
    1: 102
    2: 103
    3: 104
    ....
    646: 985
    647: 986
    648: 987
    is that how many beers are on your wall?
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  5. #5
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    Re: different digits

    No, no! DenisB would never allow that many beers to remain on his wall!
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  6. #6
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    Re: different digits

    Why is every(hic!)body always a'pic(hic!)in on me...
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  7. #7
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    Re: different digits

    i dont understand your method. i dont see why using combinations is correct
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  8. #8
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    Re: different digits

    Hey markosheehan.

    Do you understand what romsek said about fixing the first digit and then looking at the possibilities for the second and third?
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  9. #9
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    Re: different digits

    i understand it now. so there are 9 different digits for the first number (0 isnt included) for the second digit there are 9 options(0 is included) for the third option there are 8
    so 9*9*8=648
    that^ seems to be the easiest way to explain it.
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  10. #10
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    Re: different digits

    Quote Originally Posted by romsek View Post
    We're talking a 3 digit number which obviously cannot start with the digit 0

    you could look at it as the numbers w/no 0 + the numbers with 1 0

    The first term is $3! \binom{9}{3}=504$
    This is the term for the numbers with no $0$

    There are $\binom{9}{3}$ ways to choose the 3 unique digits but since we do care about the ordering, i.e. $123 \neq 321$

    we multiply by $3!$ to account for all the permutations of the $3$ digits.

    The second term is $\binom{9}{1}2! \binom{8}{1} = 144$
    This term is for numbers that contain one $0$

    $\binom{9}{1}$ is just the fact that there are 9 valid digits for the first digit in the number. $0$ is not valid.

    $\binom{8}{1}$ is just the fact that since we know one of the remaining two digits is a $0$, there are only $8$ digits to choose from for the free digit.

    But again ordering matters so we multiply by $2!$ to take into account the $2!$ permutations of these two remaining digits.
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