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Math Help - permutation combination

  1. #1
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    permutation combination

    A word is a combination of 8 letters, each either A or B. Let x and y be 2 words differing in exactly 3 places. How many words differ from each of x and y in atleast 5 places?

    can someone plz explain how to goa bt this problem??
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  2. #2
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    Hello, ramanujam!

    A word is a combination of 8 letters, each either A or B.
    Let x and y be 2 words differing in exactly 3 places.
    How many words differ from each of x and y in at least 5 places?
    If I interpret the problem correctly, the answer is EIGHT.


    x and y differ in exactly three places.
    . . Which three places? .It doesn't matter.

    Suppose they differ in their first three letters.

    They may look like this:

    \begin{array}{cccccccccc} x & = & \text{(A)} & \text{(B)} & \text{(A)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}\\<br />
 y & = & \text{(B)} & \text{(A)} & \text{(B)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}<br />
\end{array}
    . . . . . \downarrow\quad\;\;\;\downarrow\quad \;\;\;\downarrow\qquad\;\downarrow \qquad\quad\;\,\downarrow\qquad \quad\;\,\downarrow\qquad\quad\;\downarrow\qquad\q  uad\;\downarrow
    \begin{array}{cccccccccc}z & = & \text{any} & \text{any} & \text{any} & \text{(opp)} & \;\,\text{ (opp)} & \;\,\text{ (opp)} & \,\text{ (opp)} & \:\text{ (opp)} \end{array}


    The word z cannot differ from x and y in the first three places.
    . . The first three places can filled in: 2^3 = 8 ways.

    The last five letters of z must be the opposites of those in x and y.
    . . There is one way to do this.

    Therefore, there are 8 \times 1 = 8 ways that z can differ from x and y
    . . in at least five digits.

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  3. #3
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    What about this z?
    ABAAAAAA=x
    BABAAAAA=y
    AABBABBB=z
    It differs from each of x & y in at least five places.
    Was that z counted?

    P.S.
    This is a postscript.
    I know that the answer is 38.
    But I confess that I do not know how to prove it.
    Logic tells me that the answer is (8)(6).
    I do not know where the overcount of 10 comes in.
    Last edited by Plato; June 12th 2007 at 03:51 PM. Reason: a postscript
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  4. #4
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    yes plato u correct the answer is 38,but i myself m not getting that answer..
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  5. #5
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    Ramanujam, I just was over thinking this problem. Letís use Sorobanís example.
    \begin{array}{cccccccccc} x & = & \text{(A)} & \text{(B)} & \text{(A)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}\\<br />
 y & = & \text{(B)} & \text{(A)} & \text{(B)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}<br />
\end{array}
    . . . . . \downarrow\quad\;\;\;\downarrow\quad \;\;\;\downarrow\qquad\;\downarrow \qquad\quad\;\,\downarrow\qquad \quad\;\,\downarrow\qquad\quad\;\downarrow\qquad\q  uad\;\downarrow
    \begin{array}{cccccccccc}z & = & \text{any} & \text{any} & \text{any} & \text{(opp)} & \;\,\text{ (opp)} & \;\,\text{ (opp)} & \,\text{ (opp)} & \:\text{ (opp)} \end{array}

    In addition to the eight he found, there are thirty more.
    Here is how to find them. Say
    ABAAAAAA=x
    BABAAAAA=y
    Change exactly one of the red letters and four of the others in the same srting.
    For example: ABBABBBB=z.
    Now z differs from each of x & y by in at least five places.
    There are thirty ways to do that: (6)(5)=30.
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  6. #6
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    Maybe my interpretation is off . . .

    A word is a combination of 8 letters, each either A or B.
    Let x and y be two words differing in exactly 3 places.
    How many words differ from each of x and y in at least 5 places?
    I understood it to mean:

    Given an x and a y which differ in exactly 3 places,
    . . how many z's are there that differ from both x and y
    . . in at least five places?


    So we could have:

    . . \begin{array}{c}x\:=\:ABA\:XXXXX \\<br />
y \:=\:BAB\:XXXXX\end{array} .where the X's indicate matching letters

    Since the first place of x and y have both A and B,
    . . it is impossible for the first place of z to differ from both of them.
    The same is true for the second and third places.

    Hence, the only way for z to differ from both x and y
    . . is in the last five places.

    So the last five places of z is the "complement" of XXXXX.
    . . And the first three places can be any of: 2^3 = 8 three-letter "words".

    Therefore, there are 8 z's . . . for the given x and y.



    Now, if you wish to make: . \begin{Bmatrix}x \:=\:AAA\:XXXXX \\ y \:=\:BBB\:XXXXX\end{Bmatrix}

    . . then there are still 8 z's that differ from both x and y is at least 5 places.


    And that is my interpretation of the word "each".

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  7. #7
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    Quote Originally Posted by Soroban View Post
    Maybe my interpretation is off . . .
    If you have six cards in your hand, then do have at least five cards? The common understanding in mathematics is that the answer to that question is yes.
    If the letter string, z, differs from x in five places and z differs from y in six places, then z differs from each of x & y in at least five places. Isn’t that the common understanding?

    The questioner has already said that the answer is 38.

    ABAAAAAA=x
    BABAAAAA=y
    Change exactly one of the red letters and four of the others in the same srting.
    For example: ABBABBBB=z.
    Now z differs from each of x & y by in at least five places.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Ramanujam, I just was over thinking this problem. Let’s use Soroban’s example.
    \begin{array}{cccccccccc} x & = & \text{(A)} & \text{(B)} & \text{(A)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}\\<br />
 y & = & \text{(B)} & \text{(A)} & \text{(B)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)} & \text{(same)}<br />
\end{array}
    . . . . . \downarrow\quad\;\;\;\downarrow\quad \;\;\;\downarrow\qquad\;\downarrow \qquad\quad\;\,\downarrow\qquad \quad\;\,\downarrow\qquad\quad\;\downarrow\qquad\q  uad\;\downarrow
    \begin{array}{cccccccccc}z & = & \text{any} & \text{any} & \text{any} & \text{(opp)} & \;\,\text{ (opp)} & \;\,\text{ (opp)} & \,\text{ (opp)} & \:\text{ (opp)} \end{array}

    In addition to the eight he found, there are thirty more.
    Here is how to find them. Say
    ABAAAAAA=x
    BABAAAAA=y
    Change exactly one of the red letters and four of the others in the same srting.
    For example: ABBABBBB=z.
    Now z differs from each of x & y by in at least five places.
    There are thirty ways to do that: (6)(5)=30.
    thanks for the nice explanation
    i understood how u got 5 in the exp 6*5=30 but how did u get six from changing one of the red letters??
    Last edited by ramanujam; June 13th 2007 at 09:51 PM.
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  9. #9
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    hi plato,i thought more
    i u meant that u could change any of the 6 red letters..which results in the presence of 6 in the exp 6*5= 30...
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