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    Find the total number of permutations of four letters selected from the word ARRANGEMENT.

    Would be :awesome: if someone would help me out with this one.
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    (Original post by tengil)
    Find the total number of permutations of four letters selected from the word ARRANGEMENT.

    Would be :awesome: if someone would help me out with this one.
    I'm not entirely sure this is correct (I hate permutations) but...

    11!/(2! x 2! x 2! x 2!)

    You have to divide by the 2! to get rid of any repetitions.

    Would someone else please confirm whether I am right?

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    You have 11 options for the first letter, then 10, then 9, then 8. But then you have to account for the four repetitions of letters so your answer should be:

     \frac{11.10.9.8}{2!.2!.2!.2!} = 495
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    (Original post by marcusmerehay)
    You have 11 options for the first letter, then 10, then 9, then 8. But then you have to account for the four repetitions of letters so your answer should be:

     \frac{11.10.9.8}{2!.2!.2!.2!} = 495
    Sanity check: how many permutations of 4 letters from ARNGEMT are there?
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    (Original post by DFranklin)
    Sanity check: how many permutations of 4 letters from ARNGEMT are there?
    Good point. It's been far too long since I did perms and combs. :sad:
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    Unless I'm missing something, this is actually fairly hard:

    First, count the number of permutations with no repeated letters.
    Next, count the number of permutations with 1 pair of repeated letters.
    Finally, count the number of permutations with 2 pairs of repeated letters.
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    (Original post by DFranklin)
    Unless I'm missing something, this is actually fairly hard:

    First, count the number of permutations with no repeated letters.
    Next, count the number of permutations with 1 pair of repeated letters.
    Finally, count the number of permutations with 2 pairs of repeated letters.
    Tried that but I always get the wrong answer

    ARRANGEMENT, 2 Rs, 2 As, 2 N, 2 E

    Every letter different
    ARNGEMT
    7*6*5*4

    2A's and two others
    1*1*6*5*(4!/2!) <--- Not sure about this but if you 4 items where two are the same you could place them in 4!/2! different ways

    This should be the same for 2Rs, 2Ns and 2Es so

    7*6*5*4+6*5*(4!/2!)*4 = 2280

    The right answer should be 1596
 
 
 

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