+3 votes

A certain tree of order n has only vertices of degree 1 and degree 3. How many degree-3 vertices does the tree have?

(A). \frac{n-1}{2}

(B). \frac{n-2}{2}

(C). \frac{n}{2}

(D). \frac{n-1}{3}

asked Jul 12, 2019 in Discrete Maths by gbeditor (24,360 points)
reshown Jul 13, 2019 by gbeditor

2 Answers

+1 vote
Ans is B.

I solved by taking counter examples

for n = 4 no of 3 degree vertex = 1

for n = 8 no of 3 degree vertex  = 3.
answered Jul 13, 2019 by tssinghharendra711 (410 points)
an order of a tree and of a b-tree means the same?
No, we can not take a Binary tree here. Because if we take Binary tree then at least one vertex should have degree 2, which is not allowed.
So, here we should take the ternary tree.
I didn't mean binary tree @lakshmanpro20, I meant B-tree

Actually, I am confused about the term order of the tree is 'n'  what does that mean?
Yes, we can take B-tree. Because here order is given.
order represents no of vertices.
0 votes
(n-b)1 + 3b=2(n-1)[2*no. of edges in tree]

slove for b.

NOTE: edges in tree = n-1
answered Jan 2 by rajatmaheshwari2572one (5,850 points)