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# Decision 1 (D1): How do I calculate the number of possible pairings? [help me] watch

1. 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.
2. 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.

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3. Thanks, I really appreciate your help!

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