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    Hi, please can somebody help me with this question.

    "By writing out arrangements of (D1)(E1)(E2)(D2), show that there are 6 different arrangements of the letters of the word DEED?

    This seems like a basic question but i can only find 4 arrangements!?

    D1E1E2D2
    D1E2E1D2
    D2E1E2D1
    D2E2E1D1

    Thanks
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    (Original post by AlexRogers)
    Hi, please can somebody help me with this question.

    "By writing out arrangements of (D1)(E1)(E2)(D2), show that there are 6 different arrangements of the letters of the word DEED?

    This seems like a basic question but i can only find 4 arrangements!?

    D1E1E2D2
    D1E2E1D2
    D2E1E2D1
    D2E2E1D1

    Thanks
    All the arrangements you've given reflect the order DEED, which in only one arrangment of the letters.

    What about D1D2E1E2, for example.
    In fact there are 24 arrangements in all.

    The question itself seems rather convoluted if you're trying to find the number of arrangements of DEED.

    I do wonder if E1, for example, is meant to represent the first E that you come to , so E1 is always before E2. Similarly for D1,D2. In that case there would only be 6 arrangements.
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    In other words, you have treated this as a permutations question for the D and the E rather than a combinations one. There are 24 permutations of all 4 letters but you need to divide out the permutations of repeated E (2!) and repeated D (2!) giving 24/(2!2!)=6

    DEED, DEDE, DDEE, EDED, EDDE, EEDD.
 
 
 
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