1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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


    User Avatar

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Permutations/Combinations - Could someone please check my work?