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Permutations ?

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic.
    e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden.


    2. Relevant equations



    3. The attempt at a solution

    first i split the problem, first the numbers then the alfabet:

    NUMBERS:
    so N1 is the number of numbers i can use which is 10 obviously. 0-9
    K is the number of positions which is 3.

    N1!/(N1-K)=10!/(10-3)!

    ALFABET:
    so N2 is the number of alfabets which is 26, K is the number of positions 3.

    N2!/(N2-K)!=26!/(26-3)!

    The ANSWER:

    (N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!)

    is the answer correct?
     
  2. jcsd
  3. Sep 6, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    That is correct, assuming every car plate has to use 3 letters and 3 digits. You can calculate that number as a final result.
     
  4. Sep 6, 2013 #3
    thanks, does the answer exclude the permutations where the same number or digit happens in all three slots? e.g. AAA 123 or ABC 111 ?
     
  5. Sep 6, 2013 #4

    CAF123

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    Gold Member

    Yes, what you have is $$3! \cdot{10 \choose 3} \times 3! \cdot {26 \choose 3}$$ which means, in the first term you select 3 distinct objects from the group of size 10. Each choice of three objects then has 6 possible rearrangements. Similar for the other term.

    If you are allowed to use the same objects more than once, then clearly the number of possible permutations will increase.
     
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