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In a complete graph of n vertices, each vertex has a degree of $n-1$.

Let number of edges =e

we know that,

Sum of degrees$=$ $2*e$

$(n-1) +(n-1)+(n-1)+.....+(n-1)(n times)=2*e$

$n(n-1)=2*e$

$e=n(n-1)/2$
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A graph is called a complete graph ($K_n)$ if each vertex is connected to the $n-1$ remaining vertices.

For eg $K_3$ graph having 3 vertices and 3 edges.

in the same way, $K_4$ having 4 vertices and 6 edges.

 

For $K_n$ graph with $n$ vertex, $\frac{n(n-1)}{2}$ edges should be there.

So with the given options, only option B is satisfied. option B is correct.

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