$$G$$ is a simple undirected graph. Some vertices of $$G$$ are of odd degree. Add a node $$v$$ to $$G$$ and make it adjacent to each odd degree vertex of $$G$$. The resultant graph is sure to be

A binary tree with $$n>1$$ nodes has $${n_1}$$, $${n_2}$$ and $${n_3}$$ nodes of degree one, two and three respectively. The degree of a node is defined as the number of its neighbours.

$${n_3}$$ can be expressed as:

A binary tree with $$n>1$$ nodes has $${n_1}$$, $${n_2}$$ and $${n_3}$$ nodes of degree one, two and three respectively. The degree of a node is defined as the number of its neighbours.

Starting with the above tree, while there remains a node $$v$$ of degree two in the tree, add an edge between the two neighbours of $$v$$ and then remove $$v$$ from the tree. How many edges will remain at the end of the process?

$$G$$ is a graph on $$n$$ vertices and $$2n-2$$ edges. The edges of $$G$$ can be partitioned into two edge-disjoint spanning trees. Which of the following in NOT true for $$G$$?