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Thread: [SOLVED] permutations 2

  1. #1
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    [SOLVED] permutations 2

    a) How many permutations are there of the letters in the word baseball?

    my thinking $\displaystyle \frac{8}{2!2!2!}=5040$

    b) How many begin with the letter a?

    c) How many end with the letter e?
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  2. #2
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    Quote Originally Posted by william View Post
    a) How many permutations are there of the letters in the word baseball?

    my thinking $\displaystyle \frac{8}{2!2!2!}=5040$

    b) How many begin with the letter a?

    c) How many end with the letter e?

    (b) There are 2 arrangements that will begin with the letter a . Thus , 7!/(2!2!)


    (c)ONly 1 arrangement will end with e . Thus , 7!/(2!2!2!)
    Last edited by mathaddict; Feb 12th 2009 at 04:35 AM. Reason: MIstakes !
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  3. #3
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    Hello, William!

    a) How many permutations are there of the letters in the word BASEBALL?

    My thinking: $\displaystyle \frac{8}{2!2!2!}\:=\:5040$ . . . . Right!

    b) How many begin with the letter A?

    Place one of the $\displaystyle A$'s in front: .$\displaystyle A\:\_\:\_\:\_\:\_\:\_\:\_\:\_$

    Then the other seven letters $\displaystyle \{A,B,B,E,L,L,S\}$
    . . can be arranged in $\displaystyle \frac{7!}{2!2!} \:=\:1260$ ways.



    c) How many end with the letter E?

    Place the $\displaystyle E$ on the end: .$\displaystyle \_\:\_\:\_\:\_\:\_\:\_\:\_\:E$

    Then the other seven letters $\displaystyle \{A,A,B,B,L,L,S\}$
    . . can be arranged in $\displaystyle \frac{7!}{2!2!2!} \:=\:630$ ways.

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