Euler Path

Question

Which of the following Graph has Euler Path but is not an Euler Graph?
A. K_1,1 
B.K_2,10 
C.K_2,11
D.K_10,11.

Comments

Explanation in simple words with Generalized Formulae is Welcome !!!

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euler path exist but euler circuit does not exist iff the no of vertices of odd degree is exactly TWO.!
which is true in case of option (A) only :)

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euler path exist but euler circuit does not exist iff the no of vertices of odd degree is exactly TWO.!

Exactly Sir If we apply the same Logic on option C then?

There are also 2 Odd Degree vertices?

Why isn't that TRUE? 

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K_2,11 is complete bipartite graph ryt .?

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Yes, it is and two nodes are having degree 11.And that is odd , so there can be Euler Path in That K_2,11 also. That's what I am thinking. :)

Sir, is there any wrong in that logic?

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A connected graph is Euler graph iff it has at most two odd degree vertices.

Answer 1

option A is right

euler path is possible but euler cycle is not possible so it is not a euler graph

Comments

What About Option C?

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Option C is also satisfying the criteria so isn't it Eular path ?

Answer 2

The degree-sequence of a Complete Bipartite Graph $K_{p,q}$ is  $(p,p,...,p,q,...,q)$  where p is repeated q times and q is repeated p times. Therefore, $K_{2,11}$ would have a degree sequence (2,2,2,2,2,2,2,2,2,2,2,11,11). 

Since $K_{2,11}$ is a connected graph with exactly two of its vertices having odd degrees while remaining having even degrees, it will have a Euler path.

Hence options A and C would be right.

Also read,

https://math.stackexchange.com/questions/221512/hamilton-euler-circuit-path