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Math Help - String of Letters

  1. #1
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    String of Letters

    In a string of letters B U U B U U U U B, the 1st B is followed by two U's, the
    2nd B is followed by four U's, and the 3rd B is following by no U's. How many strings of 17 U's and 4 B's have every B followed by:
    1.) No U?
    2.) at most one U?
    3.) exactly one U?
    4.) at least one U?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fifthrapiers View Post
    In a string of letters B U U B U U U U B, the 1st B is followed by two U's, the
    2nd B is followed by four U's, and the 3rd B is following by no U's. How many strings of 17 U's and 4 B's have every B followed by:
    1.) No U?
    If no B is followed by a U the B's must be the last 4 chars in the string, so
    there is just one such string.

    RonL
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  3. #3
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    Quote Originally Posted by fifthrapiers View Post
    In a string of letters B U U B U U U U B, the 1st B is followed by two U's, the 2nd B is followed by four U's, and the 3rd B is following by no U's. How many strings of 17 U's and 4 B's have every B followed by:
    1.) No U?
    2.) at most one U?
    3.) exactly one U?
    4.) at least one U?
    Both #1 & #3 have the same answer. In #3 the last 8 letters are BUBUBUBU.

    In some ways #2 is the trickiest. The answer is the sum k=0 to 4 combin(4,k).

    For #4, think of the 17 Us in a line with spaces between them for one B.
    How many spaces are there before of some U ( ^U^U^U)?
    We can put each Bs into any one of those different spaces (no two in the same space) .
    How many ways?
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  4. #4
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    Quote Originally Posted by Plato View Post
    Both #1 & #3 have the same answer. In #3 the last 8 letters are BUBUBUBU.

    In some ways #2 is the trickiest. The answer is the sum k=0 to 4 combin(4,k).

    For #4, think of the 17 Us in a line with spaces between them for one B.
    How many spaces are there before of some U ( ^U^U^U)?
    We can put each Bs into any one of those different spaces (no two in the same space) .
    How many ways?
    Hmmm. I agree with #1 and #3 both being 1 string.

    For #2, however, I got this as every possible combination:

    TTTTTTAAAA
    TTTTTAAAAT
    TTTTTAAATA
    TTTTTAATAA
    TTTTTATAAA
    TTTTAAATAT
    TTTTAATATA
    TTTTATATAA
    TTTATATATA
    TTATATATAT

    So 10. Not 15 as you suggested.

    And for d, it's going to be some really, really big number. Still have no idea on this one. My thoughts:

    for d, you basically have 5 places where you have n U's

    UBUBUBUBU

    Each one of those T's can be any number of T's from 1-13, except the first one

    The left one has a min val of 0, and the right 4 have a min val of 1

    The number of possible combinations in which the sum of the 5 numbers is 17 is the answer to #4?

    Thoughts?
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  5. #5
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    2380 for #4? Trying to create a program for it but not sure how successful it'll be.
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    Hmmm. I agree with #1 and #3 both being 1 string.

    For #2, however, I got this as every possible combination:

    TTTTTTAAAA
    TTTTTAAAAT
    TTTTTAAATA
    TTTTTAATAA
    TTTTTATAAA
    TTTTAAATAT
    TTTTAATATA
    TTTTATATAA
    TTTATATATA
    TTATATATAT

    So 10. Not 15 as you suggested.
    you missed
    TTTTAATAAT
    TTTTATAAAT
    TTTTATAATA
    TTTATATAAT
    TTTAATATAT
    TTTATAATAT

    SO 16
    you have
    A-A-A-A-
    Thesefour blanks eithercontain T or not . hance there are 2^4ways.
    Keep Smiling
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    In a string of letters B U U B U U U U B, the 1st B is followed by two U's, the
    2nd B is followed by four U's, and the 3rd B is following by no U's. How many strings of 17 U's and 4 B's have every B followed by:
    1.) No U?
    2.) at most one U?
    3.) exactly one U?
    4.) at least one U?
    for d
    you have thefolloing elements to arrange
    BU,BU,BU,BU,13U
    THSE CAN BE DONE IN
    \frac{17!}{13!4!}

    KeepSmiling
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  8. #8
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    Yup! I got that answer 17!/(13!4!) by the program I created (2380 ways).

    Thanks for showing the easier way of finding it.
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