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Math Help - Combinations

  1. #1
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    Please Help

    Here is the problem:

    A sequence of letters of the form abcba is an example of a palindrome of five letters.

    a. If a letter may appear more than twice, how many palindromes of five letters are there? of six letters?

    b. Repeat part a under the condition that no letter appears more than twice.

    I do not know how to go about doing this problem. Any help would be appreciated. thanks.
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  2. #2
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    Hello, ashkash!

    I used Brute Force . . . and started making a list.


    A sequence of letters of the form ABCBA is an example of a palindrome of five letters.

    (a) If a letter may appear more than twice,
    how many palindromes of five letters are there? of six letters?

    With five letters, consider the first three letters.

    3 different letters: . ABC\;\;\Rightarrow\;\;ABCBA

    2 different letters: . \begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABABA \\ AAB\;\;\Rightarrow\;\;AABAA \\ ABB\;\;\Rightarrow\;\;ABBBA\end{array} \begin{array}{ccc}* \\ * \\ *\end{array}

    . . Only one letter: . AAA\;\;\Rightarrow\;\;AAAAA\;\;\;\:*



    With six letters, again consider the first three letters.

    3 different letters: . ABC\;\;\Rightarrow\;\;ABCCBA

    2 different letters: . \begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABAABA \\ AAB\;\;\Rightarrow\;\;AABBAA \\ ABB\;\;\Rightarrow\;\;ABBBBA\end{array} \begin{array}{ccc}* \\ * \\ * \end{array}

    . . Only one letter: . AAA\;\;\Rightarrow\;\;AAAAAA\;\;\;\:*



    (b) Repeat part (a) under the condition that no letter appears more than twice.

    Disregard answers in (a) marked with an asterisk (*).

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  3. #3
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    for the question you need to consider all 26 letters of the alphabet and not just a,b,and c.
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  4. #4
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    Quote Originally Posted by ashkash
    for the question you need to consider all 26 letters of the alphabet and not just a,b,and c.
    Would AAAAA be a valid combination?
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  5. #5
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    yes, it would.
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  6. #6
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    As soroban pointed out all you have to consider is the first three letters, each is independent of the other...

    Here's an example problem. Flip a coin. After one flip you have only two possible outcomes (2^1):

    \boxed{\begin{array}{cc}H\\T\end{array}}

    After two flips you get four possible outcomes (2^2):

    \boxed{\begin{array}{cccc}HH\\HT\\TH\\TT\end{array  }}

    After three flips you get eight possible outcomes (2^3):

    \boxed{\begin{array}{cccccccc}HHH\\HHT\\HTH\\HTT\\  THH\\THT\\TTH\\TTT\end{array}}

    You can see that after each flip the outcome multiplies by two (because a flip can give 2 different results)


    Now your problem can give 26 different results per letter, and only 3 letters matter. Therefore we get the answer of 26^3

    you get the same answer with 6 places as well
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  7. #7
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    Hello again, ashkash!

    If we are considering all 26 letters of the alphabet,
    . . we simply append a permutation factor.



    With five letters, consider the first three letters.

    3 different letters: . ABC\;\;\Rightarrow\;\;ABCBA . . . one way

    There are 26 choices for the " A",
    . . 25 choices for the " B",
    . . and 24 choices for the " C".

    Hence, there are: . 26\cdot25\cdot24 \times 1 \:=\:15,600 ways.


    2 different letters: . \begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABABA \\ AAB\;\;\Rightarrow\;\;AABAA \\ ABB\;\;\Rightarrow\;\;ABBBA\end{array} . . . three ways

    There are 26 choices for the " A"
    . . and 25 choices for the " B".

    Hence, there are: . 26\cdot25 \times 3 \:=\:1950ways.


    Only one letter: . AAA\;\;\Rightarrow\;\;AAAAA . . . one way

    There are 26 choices for the " A".

    Hence, there are: . 26 \times 1 \:=\:26 ways.


    Therefore, there are: . 15,600 + 1950 + 26 \:=\:\boxed{17,576\text{ ways.}}

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  8. #8
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    just to clear confusion: \underbrace{\overbrace{15600+1950+26}^{\text{Sorob  }\!\!\text{an's Way}}=17576=\overbrace{26^3}^{\text{my way}}}_{\text{both ways give the same answer}}
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