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What is the minimum number of edges which must be removed from a complete bipartite graph of six nodes K(6) so that the remaining graph is a planar?

**Explain with exp why your answer is Right .**

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It is $K_{3, 3}$ because other combinations result in planar graphs. To regain planarity, we need to remove only 1 edge.

Non-planarity is introduced when we try to connect two vertices which are otherwise unreachable(i.e., require intersection with other edges) *with respect to each other*.

Lets say we start with $K_{2, 3}$ which is planar. Lets name the partitions $A$(with 2 vertices) and $B$. One can observe that there is one(and only one) vertex(in B) say $h$ which is now *hidden* in the sense that it cannot be reached from any vertex in $A$. Now trying to add another vertex $n$ in A and connecting it with all vertices of B, will require connecting $n$ with $h$, which is possible only through intersection with other edge(s) and hence the introduction of *non-planarity*.

My reasoning is informal and possibly sloppy, Would anyone please provide rigorous proof/explanation for this problem.