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Hexadecimal Probability

  1. Apr 21, 2010 #1

    1. The problem statement, all variables and given/known data

    1.(a) How many 4 digit hexadecimal numbers can be formed using the hexadecimal digits 1. 5, 6, 9, A, C?(Repetition is allowed)

    (b) How many four digit hexadecimal numbers with distinct digits can be formed using the hexadecimal digits above?

    (c) What is the probability that a randomly chosen four digit hexadecimal number will have distinct digit?

    2. Relevant equations
    The number of ways of choosing r times from an n set is n!/(n-r)!

    3. The attempt at a solution

    For (a), since repetition is allowed, I did 6^4= 1296
    For (b), since repetition is not allowed, I did 6!/(6-4)! = 6*5*4*3= 360

    Now, (c) is where I start to run into problems.

    since it no longer references the above set, then I am assuming that you should take all the possible Hex digits into consideration. Therefore there are 15 possible digits in the set.

    Would you do, 15^4=50625 to get the total number of combination's possible (including repetitions) and then find out 15!/(15-4)! = 32760 to find out the possible combinations with distinct digits.

    Therefore ending up with 32760/50625?

    Please inform me if this is the correct answer,

    Thank you in advance.
  2. jcsd
  3. Apr 21, 2010 #2


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    Your method is correct, but there are 16 possible digits, 0 through F.
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