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Math Help - Counting principles, three letter sequence w/o repeated letters, containing e or f

  1. #1
    DBA
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    Counting principles, three letter sequence w/o repeated letters, containing e or f

    How many three-letter sequences without repeated letters can be made using a,b,c,d,e,f in which either e or f (or both) is used?

    I looked at the total sequences posible with all letters:
    6*5*4=120 this is w/o replacement

    The I looked at the amount of sequences without e or f:
    4*3*2=24

    Then I substract the amount of sequences without e or f from the total amount of possible sequences:
    (6*5*4)-(4*3*2)=96 amount of sequences with e or f

    But now I want to find the amount of sequences with e or f directly.
    Here I have a problem because it is e or f.

    If I would look for "without repetition and containing the letter g," I would do
    g _ _ 5 * 4 = 20
    NOg g _ 5 * 4 = 20
    NOg NOg g 5 * 4 = 20 and then add it up 20 + 20 + 20 = 60

    How can I do this with g or f?
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  2. #2
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    I think you have to use an inclusion-exclusion argument. The number of triples containing an e is 60. So is the number containing an f. But if you just add those two, you are double-counting the triples that contain both e and f. So you need to count how many there are of those (in fact, 24), and subtract that number. Then that gives you the answer 60 + 60 – 24 = 96 for the number containing either e or f (or both), in agreement with the previous method.
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  3. #3
    DBA
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    Thank you for your answer. But I was looking for a solution withour a substruction. The book gives the answer (2*3*4*3) + (3*2*4) = 96 and I do not understand how to get this answer.
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    Quote Originally Posted by DBA View Post
    Thank you for your answer. But I was looking for a solution withour a substruction. The book gives the answer (2*3*4*3) + (3*2*4) = 96 and I do not understand how to get this answer.
    If exactly one of e, f occurs then there are 2 choices for which of them it is, 3 choices for the position in which it occurs, 4 choices for the letter in the first vacant position, and 3 choices for the letter in the remaining vacant position: 2*3*4*3 = 72 possibilities.

    If both of e, f occur then there are 3 choices for the position in which e occurs, 2 choices for the position in which f occurs, and 4 choices for the letter to go in the remaining vacant position: 3*2*4 = 24 possibilities.
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