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Math Help - Probability: No. of 4-letter string formed by ...

  1. #1
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    Probability: No. of 4-letter string formed by ...

    Question

    How many no. of 4-letter strings can be formed by the letters {A,B,C,D,E,F} with no repetitions and A cannot be the first letter?



    My Answer

    I know the total no. of 4-letter strings with no repetitions is 6x5x4x3=360

    Then it comes to the next consideration: how many no. of combination with A being the first letter...

    Should it be 5x4x3=60?[/SIZE]


    Then the answer is 360-60=300

    Not sure if this is correct ...
    Last edited by pozat; March 8th 2010 at 07:36 AM. Reason: typing mistake
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  2. #2
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    Quote Originally Posted by pozat View Post
    Question

    How many no. of 4-letter strings can be formed by the letters {A,B,C,D,E,F} with no repetitions and A cannot be the first letter?



    My Answer

    I know the total no. of 4-letter strings with no repetitions is 6x5x4x3=360

    Then it comes to the next consideration: how many no. of combination with A being the first letter...

    Should it be 5x4x3=60?[/SIZE]


    Then the answer is 360-60=300

    Not sure if this is correct ...
    That is correct. It is also 5\cdot 5\cdot 4\cdot 3=300
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  3. #3
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    You got it.

    Another way of looking at it is that the first letter can be anything except for "A"--there are 5 possibilities for the first letter. The second letter can be anything except for whatever gets used for the first letter. For example, if the first letter were "D", then the second letter can be anything in {A,B,C,E,F}, so that's 5 more. Then there are 4 possibilities for the third letter, and three possibilities for the fourth letter. That's a total of 5*5*4*3=300 possible 4-letter strings using {A,B,C,D,E,F} without repititions.
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  4. #4
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    The answer is correct. Thanks to ALL.
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