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Math Help - Arrangements of letters in COONABARABRAN

  1. #1
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    Arrangements of letters in COONABARABRAN

    Find the number of different arrangements of the letters in the word
    COONABARABRAN
    How many of those arrangements (i) start with the six vowels; (ii) contain the
    four letters A consecutively; (iii) contain no two consecutive letters A?


    Answer:

    Total number of arrangements: \binom{13}{1,2,2,2,4} = 16216200

    (i) Put As and Os at start \binom{6}{2,4} = 15. Then number of arrangements of NNRRBBC is \binom{7}{2,2,2} = 630. 15 * 630 = 9450


    (ii) Treat AAAA as one item. That means there are 10 items to arrange - AAAA C O O N N B B R R. \binom{10}{2,2,2,4} = 18900 ways.


    (iii) _C_O_O_N_N_B_B_R_R_ ---> Put As in the gaps. This can be done \binom{10}{4} = 210 ways. Number of arrangements of COONNBBRR = \binom{9}{2,2,2,2} = 22680. 210 * 22680 = 4762800.


    Are those answers correct?
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  2. #2
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    Re: Arrangements of letters in COONABARABRAN

    Quote Originally Posted by nukenuts View Post
    Find the number of different arrangements of the letters in the word COONABARABRAN
    How many of those arrangements (i) start with the six vowels; (ii) contain the
    four letters A consecutively; (iii) contain no two consecutive letters A?
    (ii) Treat AAAA as one item. That means there are 10 items to arrange - AAAA C O O N N B B R R. \binom{10}{2,2,2,{\color{red}4}} = 18900 ways.
    In (ii) there is no need for the 4, "AAAA" is only one.
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  3. #3
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    Re: Arrangements of letters in COONABARABRAN

    Quote Originally Posted by Plato View Post
    In (ii) there is no need for the 4, "AAAA" is only one.
    Oops, meant to treat it as one item but wrote it down wrong.
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