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    First, is it considered a chinese postman method when there are more than 2 odd vertices or is it called the route inspection algorithm?

    And, is it still considered semi-eulerian with 4 odd vertices? Or is it only 2?

    Last question, in the route inspection algorithm, the edges with odd degrees must be paired up.
    2 odd vertices = 1 possible pairing.
    4 odd verticies = 3 possible pairings.
    6 odd verticies = 15 pairings.
    8 odd vertices = 105 vertices.

    How do I calculate the number of possible pairings? What is the formula?

    Thank you and I appreciate your help.
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    To be semi-Eulerian there must be 2 nodes with odd order.

    You don't need to know how many pairings there are. It's a natural question to ask though.

    Spoiler:
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    \displaystyle \frac{n!}{2^{n/2} (n/2)!}}
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    Thanks, I really appreciate your help!
 
 
 
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Updated: January 4, 2013
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