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# DM-Graphs-Q4

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+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 1 flag
reshown Jul 13, 2019

## 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 (450 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
https://en.wikipedia.org/wiki/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.
I am getting answer as (n-1)/3 when I take random values of n with deg 1 and deg 3 vertices
you are doing the mistake of taking degree 4 as 3. for example
.
/ | \
.
/ | \
here only the top most element has degree 3, bottom left vertex has degree 4 if you see carefully. I was doing the same mistake initially and I also got option D by taking random cases.
+1 vote
(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, 2020 by (8,040 points)
Answer:

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