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Counting and probablity addition rule?

  1. Nov 4, 2009 #1
    A calculator has an eight -digit display and a decimal point that is located at the extreme right of the number displayed, at the extreme left or between any pair of digits. The calculator can also display a minus sign at the etreme left of the number. How many distinct numbers can the calculator display? (not the certain numbers are equal, such as 1.9, 1.90, 01.900, and should, therefore, not be counted twice.)

    i have
    12 choices for the first digit , 10 for the 2nd, 9 for the 3rd, 8 for the 4th, 7 for the 5th, 6 for the 6th, 5 for the 7th, 4 for the 8th.

    But this is using the multiplication rule and doesn't account for the numbers such as 1.9, 1.90, 01.900 which the addition rule would. I can't figure out else to do it and am not sure if i'm supposed to use the addition rule.
  2. jcsd
  3. Nov 4, 2009 #2


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    Welcome to PF!

    Hi Tleeves! Welcome to PF! :smile:
    uhh? except for the repetitions, it would be 10 choices for the first digit , 10 for the 2nd, 10 for the 3rd, … 10 for the 8th.

    I think you'll need to deal with each "length" separately (where the length of a number would be the length from the first non-zero digit to the last non-zero digit), to avoid double-counting the repetitions.
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