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Math Help - probability

  1. #1
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    probability

    Find how many distinct number greater than 5000 and divisible by 3 can be formed from digits 3,4,5,6,and 0. Each digit being used at most once in any number.
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  2. #2
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    Re: probability

    Hello, Trefoil2727!

    Find how many distinct numbers greater than 5000 and divisible by 3 can be formed
    from digits 3,4,5,6,and 0, each digit being used at most once in any number.

    A number is divisible by 3 if its sum of digits is divisible by 3.


    Four-digit numbers

    The number begins with 5: . 5\,\_\,\_\,\_
    The other 3 digits can be: . \{0,3,4\},\:\{0,4,6\},\:\{3,4,6\}
    . . Each has 3! permutations.
    Hence, there are: 3\cdot3! \,=\,18 four-digit numbers that begin with 5.

    The number begins with 6: . 6\,\_\,\_\,\_
    The other 3 digits can be: . \{0,4,5\},\:\{3,4,5\}
    . . Each has 3! permutations.
    Hence, there are: 2\cdot3! \,=\,12 four-digit numbers that begin with 6.

    There are: 18 + 12 \,=\,{\color{blue}30} four-digit numbers.


    Five-digit numbers

    The digits \{0,3,4,5,6\} add up to 18, a multiple of 3.
    Hence, any five-digit number will be divisible by 3.

    There are 4 choices for the first digit.
    The other 4 digits can be permuted in 4! ways.
    Hence, there are: 4\cdot4! \,=\,{\color{blue}96} five-digit numbers.


    Therefore, there are : 30 + 96 \,=\,{\color{blue}126} such numbers.
    Thanks from Trefoil2727
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