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Someone please solve it.

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There are two cases possible 

case:1(when noof vertices is odd)

If the noof vertices (with  degree of each vertex as 1) is odd then the graph is not even possible because the sum of degree of all the vertices is odd ( but using hand shaking lemma you can say that sum of degree is always even .) 

 

Case :2 (when noof vertices is even and all the vertices ga e degree 1) 

Using handshaking lemma you can find the noof edges . 

hand shaking lemma:  sum of degree of all verties = 2*(noof edges)

So from above question(option B) sum of degree of all vertices = 10  10=2*edges. Therefore e=5(noof edges)

But you know that a graph with n vertices to be connected the minimum noof edges is n-1 . So with n=10 the minimum edges is 9 but we got just 5 in option B . So option B was not possible . All the remaining are possible . 

NOTE ::

see the below link if you want to know , For a connected graph with n vertices the minimum noof edges is n-1 . 

https://stackoverflow.com/a/42600711

 

 

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