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6 April, 20:25

How many 5 digit numbers are there whose digits sum to 39?

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  1. 6 April, 21:53
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    The distribution of the possible digits of the numbers are

    1.) 9, 9, 9, 9, 3 [Number of arrangements = 5! / 4! = 120 / 24 = 5]

    2.) 9, 9, 9, 8, 4 [Number of arrangements = 5! / 3! = 120 / 6 = 20]

    3.) 9, 9, 9, 7, 5 [Number of arrangements = 5! / 3! = 120 / 6 = 20]

    4.) 9, 9, 9, 6, 6 [Number of arrangements = 5! / (3! x 2!) = 120 / 12 = 10]

    5.) 9, 9, 8, 8, 5 [Number of arrangements = 5! / (2! x 2!) = 120 / 4 = 30]

    6.) 9, 9, 8, 7, 6 [Number of arrangements = 5! / 2! = 120 / 2 = 60]

    7.) 9, 9, 7, 7, 7 [Number of arrangements = 5! / (3! x 2!) = 120 / 12 = 10]

    8.) 9, 8, 8. 8, 6 [Number of arrangements = 5! / 3! = 120 / 6 = 20]

    9.) 9, 8, 8, 7, 7 [Number of arrangements = 5! / (2! x 2!) = 120 / 4 = 30]

    10.) 8, 8, 8, 8, 7 [Number of arrangements = 5! / 4! = 120 / 24 = 5]

    Number of 5 digit numbers whose digit sum up to 39 = 5 + 20 + 20 + 10 + 30 + 60 + 10 + 20 + 30 + 5 = 210
  2. 6 April, 22:42
    0
    I would have to say 15 total ways
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