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Math Help - Help with combinatorics problem

  1. #1
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    Help with combinatorics problem

    Hi

    I would appreciate if someone could explain how to solve the following problem:


    "How many different five letter "words" can you write with the letters BILBANA"


    The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:

    A+B = 5!
    A+A = 5!/2!
    B+B = 5!/2!
    A+B+B = ?
    A+A+B = ?
    A+A+B+B = ?



    Thanks in advance
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  2. #2
    MHF Contributor
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    Hello Drdumbom
    Quote Originally Posted by Drdumbom View Post
    Hi

    I would appreciate if someone could explain how to solve the following problem:


    "How many different five letter "words" can you write with the letters BILBANA"


    The answer is 690 but I'm not entirely sure how to get there. I divide the problem into different cases:

    A+B = 5!
    A+A = 5!/2!
    B+B = 5!/2!
    A+B+B = ?
    A+A+B = ?
    A+A+B+B = ?



    Thanks in advance
    You're right so far. Let me continue:

    With A + B + B there will be two other letters chosen from the remaining 3. So we have two processes:

    • Choose 2 out of 3 remaining letters. This can be done in {^3C_2}=3 ways.


    • Arrange the 5 letters, which include a repeated B. This can be done in \frac{5!}{2!}= 60 ways.

    So A + B + B can be done in 3\times 60 =180 ways.

    Similarly A + A + B can be done in 180 ways.

    And then in a similar way A + A + B + B + one other letter can be chosen and arranged in {^3C_1}\times\frac{5!}{2!2!}=90 ways.

    Therefore the total no of 'words' = 120 + 60 + 60 + 180 + 180 + 90 = 690.

    Grandad
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