DSSSB PGT 2021
October 12, 2023
Nielit Scientist-D 2016 march
October 12, 2023
DSSSB PGT 2021
October 12, 2023
Nielit Scientist-D 2016 march
October 12, 2023

UGC NET CS 2014 Dec-Paper-2

Question 2
A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph ?
A
5
B
n-3
C
20
D
11
Question 2 Explanation: 
Method-1:
Here, they are clearly mentioned that “certain tree”
The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.
→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.
→ Two vertices of degree 4, we can write into (2*4)
→ One vertex of degree 3, we can write into (1*3)
→ One vertex of degree 2, we can write into (1*2).
→ Other vertices(X) have degree 1, we can write into (X*1)
Step-1: (2*4)+(1*3)+(1*2)+(X*1)
=2(X+4-1) [Note: ‘X’ value is not given ]
X=7
Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.
=X+4
=7+4
=11
Method-2:
Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1
D degree = 2
A degree = 4
K degree = 4
G degree = 3
Correct Answer: D
Question 2 Explanation: 
Method-1:
Here, they are clearly mentioned that “certain tree”
The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.
→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.
→ Two vertices of degree 4, we can write into (2*4)
→ One vertex of degree 3, we can write into (1*3)
→ One vertex of degree 2, we can write into (1*2).
→ Other vertices(X) have degree 1, we can write into (X*1)
Step-1: (2*4)+(1*3)+(1*2)+(X*1)
=2(X+4-1) [Note: ‘X’ value is not given ]
X=7
Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.
=X+4
=7+4
=11
Method-2:
Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1
D degree = 2
A degree = 4
K degree = 4
G degree = 3
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