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A string is an ordering of letters ILLINOIS a

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data
    In this problem a string is an ordering of letters ILLINOIS
    a. how many such strings are there
    b. in how many strings, such as LSLOINII, does each L appear before each I?
    c. In how many strings does one I appear the L?


    3. The attempt at a solution
    a. this is pretty easy, just using partition formula
    [itex]\frac{8!}{2!3!1!1!1!}[/itex]
    which is 3360

    however, for question 2 and 3 im not so understand about it
    for question 2, considering each L appear before each I, should i conclude them as a partition?
    which eventually, leads to 8C5 [itex]\times[/itex] 3C1 [itex]\times[/itex] 2C1 [itex]\times[/itex] 1C1
    need some hints for question 3
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 23, 2011 #2
    Re: Combinatorics

    For part b. I'd go like this:
    Imagine for each string you highlight the Ls and Is.
    Eg. ILLINOIS --> ILLII
    How many possible dispositions you might get ?
    They are [tex]\frac{5!}{2!3!}= 10[/tex]
    only one of them is the "good" one (LLIII) so, I'd say only 1 out of 10 of all the possible original combinations is good. That is 3360/10 = 336

    Part c.
    I assume the correct question is "In how many strings does one I appear before the L?"

    The reasoning is pretty the same as before you just have one I stuck at the beginning
    ILLII
    Then you have to find in how many ways you can reaggange the four remaining letters: LLII
    which is similarly as before [tex]\frac{4!}{2!2!}=6[/tex]
    So I'd say only 1 out of 6 of the original arrangements (560).

    I just tried as you did, don't take my solutions for sure.
     
  4. Aug 23, 2011 #3
    Re: Combinatorics

    wow, you open another door for me to see about it, because typically, i still struggle in the ways to determining the correct ways to find the actual answer
    thx for teaching me such a nice approach to these problems
     
    Last edited: Aug 23, 2011
  5. Aug 24, 2011 #4
    Re: Combinatorics

    In part c. there's a mistake:
    it must be read 6 out of 10 of all possible original rearrangements:
    3360*(6/10) = 2016
     
  6. Aug 24, 2011 #5
    Re: Combinatorics

    hmmm, why is it required to be assuming there must be 6 out of 10? i dun get the statement here
     
  7. Aug 24, 2011 #6
    Re: Combinatorics

    Well, we can count them, there's not a lot of them.

    L: 0
    I: 1

    1) 00111
    2) 01011
    3) 01101
    4) 01110
    5) 10011
    6) 10101
    7) 10110
    8) 11001
    9) 11010
    10) 11100
     
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