retagged by
18,105 views

9 Answers

Best answer
76 votes
76 votes
Let $m$ be minimum degree and $M$ be maximum degree of a graph, then $\color{navy}{m \le \frac{2E}{V} \le M}$

$m = 3, E = 25, V = ...?$

So, $3 \le \frac{2*25}{V}$

$V\le \frac{50}{3}$

$V \le 16.667 \color{maroon}{\Rightarrow V = 16}$
edited by
10 votes
10 votes
k.V<=2E

so,3V<=2*25=$\left \lfloor 50/3 \right \rfloor =16$

Ans:16
8 votes
8 votes
According to Handshaking Lemma: The Sum of degree of all the vertices is equal to the twice the number of edges.

in our question:

Number of vertices = n

Number of edges = 25

3 * n = 2  * 25

3n=50

n=16.667

n =16 which is the maximum possible value
reshown by
Answer:

Related questions

1.2k
views
0 answers
0 votes
admin asked Mar 30, 2020
1,236 views
Which of the following statements is/are TRUE for an undirected graph?Number of odd degree vertices is evenSum of degrees of all vertices is evenP onlyQ onlyBoth P and QNeither P nor Q
592
views
1 answers
0 votes
admin asked Apr 1, 2020
592 views
The number of the edges in a regular graph of degree $’d’$ and $’n’$ vertices is Maximum of $n,d$n+d$nd$nd/2$
10.0k
views
1 answers
0 votes
admin asked Mar 30, 2020
10,030 views
Given an undirected graph $G$ with $V$ vertices and $E$ edges, the sum of the degrees of all vertices is$E$2E$V$2V$