The number of vertices of odd degree in any graph of ‘n’ vertices with ‘e’ edges is:

Correct option is B

This follows directly from the Handshaking Lemma in graph theory.
The Handshaking Lemma states that:
​$$\sum_{v\in V}\deg(v)=2e$$​​
where e is the number of edges.
Since 2e is always even, the sum of the degrees of all vertices must be even.
Now, consider the contribution of vertices:
· An even-degree vertex contributes an even number.
· An odd-degree vertex contributes an odd number.
The sum of an odd number of odd integers is odd, whereas the sum of an even number of odd integers is even.
Since the total sum of degrees is even, the number of vertices having odd degree must be even.
Therefore, Number of odd-degree vertices is always even.
Example
Suppose the degrees of five vertices are:
2, 3, 4, 1, 2
The odd-degree vertices are those with degrees 3 and 1.
Number of odd-degree vertices:
2
which is even.
Information Booster
1. Handshaking Lemma:
​$$\boxed{\sum\deg(v)=2e}$$​​
2. Every edge contributes 2 to the total degree count—one for each endpoint.
3. Therefore, the sum of all vertex degrees is always even.
4. Consequently, the number of vertices having odd degree is always even.
5. This result applies to every undirected graph, regardless of the number of vertices or edges.
Additional Knowledge
Why the Other Options Are Incorrect?
· (a) Odd — The number of odd-degree vertices cannot be odd.
· (b) Even — Guaranteed by the Handshaking Lemma.
· (c) n−e — There is no such general relationship.
· (d) Prime — The number may be 0, 2, 4,…, and these are not necessarily prime.