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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
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let G be a graph with 100! vertices, with each vertex labelled by a di...
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
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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)109b)110c)119d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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