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if only digits 0,1,2,3,4,5,6, and 7 may be used find number of possibilites four digit numbers

  • math -

    n!/r!(n-r)!
    8!/4!(8-4)!
    = 70

  • math -

    Ok, Andrew
    in my last answer to the same question I somehow missed the 7 and saw only 6 numbers.
    (Have to concentrate more on reading the question more carefully)


    In each case we can assume that no whole number starts with a zero, and no repeating digits for all 3 cases.
    We have 8 digits, but can only use 7 in the first position ...

    number of 4 digits numbers = 7x7x6x5 = 1470
    number of 3 digit odd numbers, (must end in an odd) = 6x6x4 = 144

    number of 3 digit numbers = 7x7x6 = 294

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