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Math Help - Combination Question

  1. #1
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    Combination Question

    I am trying to figure out the number of possible combinations which I can arrange this sequence:

    123456
    where:
    4 can also be X
    5 can also be Y
    6 can also be Z

    For example, 1 combination can be:
    12345Z

    Is the proper equation combinations = (n^n) -1
    where n is the number of variables that can be change?

    For my example I think think the number of combinations would be 8
    (3^3) -1

    Thanks for any help!
    Mr. EKO
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by MrEKO View Post
    I am trying to figure out the number of possible combinations which I can arrange this sequence:

    123456
    where:
    4 can also be X
    5 can also be Y
    6 can also be Z

    For example, 1 combination can be:
    12345Z

    Is the proper equation combinations = (n^n) -1
    where n is the number of variables that can be change?

    For my example I think think the number of combinations would be 8
    (3^3) -1

    Thanks for any help!
    Mr. EKO
    You will need to explain this more clearly.

    Is 231ZX a permitted combination for example?

    RonL
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  3. #3
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    No, they must be in order

    12345Z
    1234Y6
    1234YZ

    etc.

    Thank you.
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  4. #4
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    Quote Originally Posted by MrEKO View Post
    I am trying to figure out the number of possible combinations which I can arrange this sequence:

    123456
    where:
    4 can also be X
    5 can also be Y
    6 can also be Z
    I understand the question as,
    123 _ _ _
    Where it must be starting as 123.
    And for the blank ones we can have,
    (4,X) then (5,Y) then (6,Z)
    For each space we have 2 possibilities.
    Thus in total we have,
    (2)(2)(2)=8
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  5. #5
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    Thank you.

    I assume if I had four spots where it could be changed to a letter it would be
    (2)(2)(2)(2) = 16 possible combinations?
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  6. #6
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    Quote Originally Posted by MrEKO View Post
    Thank you.

    I assume if I had four spots where it could be changed to a letter it would be
    (2)(2)(2)(2) = 16 possible combinations?
    Yes.
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