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GATE | GATE CS 2018 | Question 56
• Difficulty Level : Medium
• Last Updated : 09 Mar, 2018

Let G be a graph with 100! vertices, with each vertex labelled by a distinct permutation of the numbers 1, 2, …, 100. There is an edge between vertices u and v if and only if the label of u can be obtained by swapping two adjacent numbers in the label of v. Let y denote the degree of a vertex in G, and z denote the number of connected components in G. Then y + 10z = _______ .

Note –This was Numerical Type question.

(A) 109
(B) 110
(C) 119
(D) None of these

Answer: (A)

Explanation: There is an edge between vertices u and v iff the label of u can be obtained by swapping two adjacent numbers in the label of v.
Then the set of swapping numbers will be {(1, 2), (2, 3), ………..(9, 9)}
There will be 99 such sets, i.e. number of edges = 99
and each vertex will have 99 edges corresponding to it.

Say graph with 3! vertices, then vertices will be like {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}…
Let’s pick vertex {123}, degree will be 2 since it will be connected with two other vertices {213} and {132}.

We can conclude that for n, degree will be n-1.

SO, degree of each vertex = 99 (as said y)
As the vertices are connected together, the number of connected components formed will be 1 (as said z).

` y+10z = 99+10(1) = 109` My Personal Notes arrow_drop_up