Question

suppose that a tree T has N vertices of degree 1,
2 vertices of degree 2, 4 vertices of degree 3,
3 vertices
of degree 4. Find N​

Answer 1

Answer:

Suppose that there are fewer than two vertices of degree one. So we can split into two cases.

Case one: there is no vertex of degree one. Since we know that every tree has n−1 edges then the total degree of any tree have to be 2(n−1). But for this case since no vertex have degree 1 then every vertex have at least a degree of 2 and since there are n vertices, the total degree is ≥2n which is a contradiction.

Case two: there is only one vertex of degree one. Similarly the total degree of any tree have to be 2(n−1). Then there are (n−1) vertices with which have degree of ≥2 while only one vertex with degree of one. Thus summing up to find the total of the vertices, we have that the total degree of vertices is ≥2(n−1)+1=2n−1 which is also a contradiction.

This thus proves the statement for both cases.

Answer 2

Answer:

The tree has 28 vertices.

Step-by-step explanation:

In any tree, the sum of the degrees of all vertices is equal to twice the number of edges.

Therefore, we have:

1N + 2(2) + 3(4) + 4(3) = 2E

Simplifying, we get:

N + 2 + 12 + 12 = 2E

N + 26 = 2E

Since the tree has N vertices, it has N-1 edges.

Therefore:

2E = 2(N-1) = 2N - 2

Substituting this into the previous equation, we get:

N + 26 = 2N - 2

28 = N

Therefore, the tree has 28 vertices.

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