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Math Help - Combinatorics QUESTION

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    Combinatorics QUESTION

    Let A = {a_1, a_2, . . . , a_k} be an alphabet, and let n_i denote the number of appearances of
    letter a_i in a word. How many words of length n in the alphabet A are there for which
    k = 3, n = 10, n_1 = n_2 + n_3 and n_2 is even?
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    Re: Combinatorics QUESTION

    this is not calculus ... thread moved to Discrete Math
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    Re: Combinatorics QUESTION

    Quote Originally Posted by maximus101 View Post
    Let A = {a_1, a_2, . . . , a_k} be an alphabet, and let n_i denote the number of appearances of
    letter a_i in a word. How many words of length n in the alphabet A are there for which
    k = 3, n = 10, n_1 = n_2 + n_3 and n_2 is even?
    "let n_i denote the number of appearances of letter a_i in a word"? Which word? Are you saying that there's a single unknown-yet-fixed word in the alphabet A which defined the numbers n_1, n_2, n_3,... , n_k ?
    OR, are you treating n_1, n_2, n_3,... , n_k as k distinct functions functions from the domain of all possible finite words in A into the non-negative integers?
    I would guess the later, but I'm not sure.

    With that assumption:

    10 = n_1+n_2 + n_3 = (n_2+n_3) + n_2 + n_3 = 2(n_2 + n_3).

    \text{Thus } n_3 = 5-n_2 \text{ and } n_1 = n_2 + n_3 = 5.

    \text{Case }n_2 = 0:
    \text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 0 = 5.

    \text{Case }n_2 = 2:
    \text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 2 = 3.

    \text{Case }n_2 = 4:
    \text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 4 = 1.

    \text{So ANSWER } = \text{ number of ways} (n_1 = 5, n_2 = 0, n_3 = 5)

    + \text{number of ways}(n_1 = 5, n_2 = 2, n_3 = 3)

    + \text{number of ways}(n_1 = 5, n_2 = 4, n_3 = 1),

    \text{and each term there is a clearer combinatorial problem.}

    \text{So ANSWER } = \text{ number of distinguishable permutations of } "AAAAACCCCC"

    + \text{ number of distinguishable permutations of } "AAAAABBCCC"

    + \text{ number of distinguishable permutations of } "AAAAABBBBC".

    For the middle example: imagine there are 10 slots, numbered 1 through 10, that you need to fill with those letters to make your word. Choose 5 slots to fill with the A's, and then, from the 5 remaining slots, choose 2 to fill with the B's (Obviously the rest are then filled in with C's.) The number of ways to do that will be the number of distinguishable length 10 words using 5 A's, 2 B's and 3 C's.
    Last edited by johnsomeone; October 21st 2012 at 04:38 PM.
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