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Math Help - Counting

  1. #1
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    Nov 2006
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    Counting

    1.) Suppose we have a string that consists of 0's and 1's, & that it is broken up after every 1 & after every 0. Then, the resulting fragments (note that they are NOT necessarily in order) will be the following:

    1-fragments: 0, 001, 01, 01
    0-fragments: 0, 10, 0, 10, 10.

    a.) How many strings exist with the 1-fragments
    b.) (Same as above but with 0-fragments)
    c.) Determine ALL strings that have the 1-fragments and the 0-fragments.


    2.) Is it possible for a string to be ambigiuous if its broken up in to its 0-fragments and 1-fragments? Give an example if it can be, or else explain why it cannot be.



    My work:

    For #1a, I said that the answer is 3, and I got it by:

    3!/(2!*1!) = 3;

    #1b, I said it was 5! = 5*4*3*2*1 = 120

    and #1c, which I think I did incorrectly:

    001010100,
    010010100
    010100100

    #2...no idea.
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  2. #2
    Junior Member F.A.P's Avatar
    Joined
    Dec 2006
    Posts
    26
    Quote Originally Posted by Bloden View Post
    1.) Suppose we have a string that consists of 0's and 1's, & that it is broken up after every 1 & after every 0. Then, the resulting fragments (note that they are NOT necessarily in order) will be the following:

    1-fragments: 0, 001, 01, 01
    0-fragments: 0, 10, 0, 10, 10.
    (a) We have four "letters" for the 1-fragments. The zero must be the last value. The number of distinct "words" we can write with these letters is

    \frac{3!}{1!2!} = 3

    (b) We have that only two of the 5 fragments being distinct, appearing 2 and 3 times each.

    \frac{5!}{2!3!} = 10

    (c) We have three strings to evaluate from (a)

    00101010 with 0-fragments 0, 0, 10, 10, 10

    01001010 with 0-fragments 0, 100, 10, 10

    01010010 with 0-fragments 0, 10, 100, 10

    Only one, the first string, has the given 1-fragments and the 0-fragments.
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