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Permutations/Combinations - Could someone please check my work?

  1. Nov 24, 2005 #1
    Hey everyone,

    I have an assignment on Permutations and Combinations and was wondering if some of you could go over my work and ensure that I have this aced. I really need to get my math mark up for University admissions so I need to ace this assignment.

    Thanks a lot for your help!

    1) In how many distinguishable ways can the letters of the word MISSISSIPPI be arranged?

    Number of distinguishable arrangements = 11!/(4!4!2!) = 34,650

    2) Find the number of ways of arranging all the letters AAABBBCCDD in a row so that no two A's are side-by-side.

    Number of arrangements = 7!/(3!3!2!2!) = 35

    what I did: I paired the letter A up with another letter, by viewing them as 1 letter no 2 A's would ever be side by side. This makes 7 letters, and then I divided out the repeats.

    3) How many distinguishable four letter arrangements can be made from teh letters of the word MISSISSIPPI?

    Number of four letter arrangements = (ways with no repeats) + (ways with all 4 letters repeated) + (ways with 2 doubles) + (ways with a triple and a single)

    = (4c4)x4 + (2c1) + (3c2)x(4!/2!2!) + (2c1)(1c1)x(4!/3!)
    = 24 + 2 + 18 + 8
    = 52

    *****If it is unclear what I have done hear please ask me*****

    Those are the first 3 questions out of 5. I want to make sure I'm on the right track before I finish the next 2, as they are significantly more difficult.
  2. jcsd
  3. Nov 24, 2005 #2
    I don't need to know the exact answers, just if i'm correct or not.

    Last edited: Nov 24, 2005
  4. Nov 26, 2005 #3


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    You got #1 right. It is indeed (11!)/(4!*4!*2!) = 34,650.

    For #2:

    Step one: calculate all possible ways taking into account repetitions: 10!/(3!*3!*2!*2!)

    Step two: consider AAA as one letter and thus calculate in how many varinats they are to gether (8!/(3!*2!*2!)

    Step three: consider AA as one letter and calculate in how many cases they are together: (9!/(3!2!2!)

    Step four: substract step 3 and step 2 from step one

    For #3 I'm still thinking. The only solution I can see right now is to count it manually. But meh, that isn't a good approach to combinatorics. Allow me a few more minutes to think of a better solution. Will post again soon. o:)
    Last edited: Nov 26, 2005
  5. Nov 26, 2005 #4


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    Okay, for #3, there can be many cases (possibilities-which letters we choose, not their arrangements in the 4 letter word):

    (a) All four letters are different (MISP) 4*3*2*1=24

    (b) If two letters repeat (2I, 2P or 2S) AND there is M

    There are 6 possibilities (M2IP/S; M2PI/S; M2SP/I).
    Therefore: 6*([4*3*2*1]/2)=72

    (c) If 2 and 2 letters repeat then there are 3 possibilities (2I2P; 2I2S; 2P2S)
    Therefore: 3*([4*3*2*1]/4)=18

    (d) If 3 letters repeat there are 3 possibilities (3SM; 3SP; 3SI)
    Therefore 4*3=12

    Therefore we obtain: 24+72+18+12=126

    I hope you got the method. o:)
  6. Nov 28, 2005 #5
    Is this correct? Can anybody verify please?
  7. Nov 28, 2005 #6
    Ok, that is what I was doing in my first post, so far so good.:tongue2:

    Good, I had that.

    Ooooh...kk i forgot this. THANKS IB!

    EDIT: I just worked this part out myself, and got a different answer for this case.

    There are 3 sets of letters to choose doubles from, and we want 1 of them to be chosen. Then we want 2 singles from the 3 sets of singles that are left to be chosen from. (ex, if 2 I's are chosen, we need 2 singles, fro meither S, P or M.) Then we have to multiply by 4! (for different positions) and divide by 2! (the repeats).

    Thus we get (3c1)*(3c2)*(4!/2!) = 108 ways.

    Alright, I had this.

    Hmmm....I don't think this is correct. I had the same idea, and now that I think of it I think we are both wrong.

    for each of the triples that can occur, there has to be another letter present. This means that if there are 3 I's, then the fourth can be an S, M or P. And, they can be in four different spots. I's and S's are the only letters that can give tripples. For each of them, there is a fourth letter needed, one of 3 letters possible. Then, for each of these arrangements there are 4! ways of arranging the letters. Thus I think the answer for this case is (2c1)*(3c1)*4!

    Where (2c1) means "two choose one" (combinations).

    So there should be 144 in this case....wierd, that can't be.

    Can anyone help?

    EDIT: I forgot to divide that answer by 3! to get rid of the repeats. This means that there are 24 ways for this case.

    IB, I think you also forgot to include the arrangements where there are only repeats, such as (IIII) or (SSSS). Thanks for your help though, I appreciate it. Without it I would never have seen my errors.

    Well, I redid it, and I get a total of 176 ways.

    Who is right? or are we both wrong?
    Last edited: Nov 28, 2005
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