A tree has \(2𝑛\) vertices of degree 1,\( 3𝑛\) vertices of degree 2, \(𝑛\)…

2019

A tree hasΒ \(2𝑛\)Β vertices of degreeΒ 1,\(Β 3𝑛\)Β vertices of degreeΒ 2,Β \(𝑛\)Β vertices of degreeΒ 3. Determine the number of vertices and edges in tree.

  1. A.

    12,11

  2. B.

    11,12

  3. C.

    10,11

  4. D.

    9,10

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Correct answer: A

Step 1: Count vertices.

Total vertices V = 2n + 3n + n = 6n.

Step 2: Sum of degrees and edges.

Sum of degrees = 1Β·(2n) + 2Β·(3n) + 3Β·n = 11n. By the handshake lemma, 2E = 11n, so E = 11n/2.

Step 3: Use the tree property.

  • For a tree, E = V βˆ’ 1, so E = 6n βˆ’ 1.

  • Equate the two expressions for E: 11n/2 = 6n βˆ’ 1.

  • Solve: multiply both sides by 2 β†’ 11n = 12n βˆ’ 2 β‡’ n = 2.

Conclusion: With n = 2, V = 6n = 12 and E = V βˆ’ 1 = 11.

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