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Let G be a graph with 10 vertices and 31 edges. If G has 3 vertices of degree 10, 1 vertex of degree 8 and 2 vertices of degree 5 and the other four vertices of degree at least 3, how many vertices are of degree 3________?

my solution:

Σ deg(v) = 2|E|

3*10 + 1*8 + 2*5 + 4*(>=3) = 2*31

4*(>=3) = 62- (30+8+10) = 14

I think we can have 3 vertices each of degree 3 and vertex of degree 5, so my answer is 3 but given answer is 2.

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3 vertex of degree 10 means it is a multigraph...in simple graph degree cannot be greater than 9...or this que may be framed wrongly
multigraph also follows this property:
Σ deg(v) = 2|E|

what's wrong?

ohk,

sum of degree  = 2 * edges

3 * 10 +  1 * 8 + 2 * 5  + x1 + x2 + x3  + x4 =  2 * 31

x1 + x2 + x3 + x4 =   62 - 48

x1 + x2 + x3 + x4 = 14

each node have degree atleast three means

x1 + x2 + x3 + x4 =   14 - 4 * 3 = 2

x1 + x2 + x3 + x4 = 2

Means either one node take 2 value , means  one node have degree 3 + 2 :  5 ( Not possible , because their are only 2 nodes of degree 5 )

Or

two nodes have 1 , 1 value       means two nodes have 4 degree , and left two nodes have 3 degree.

Hence nodes having degree  = 3  are 2.
answered by Loyal (2.9k points)
selected
( Not possible , because their are only 2 nodes of degree 5 )  i was thinking the same later because it is mentioned in the question that there 2 vertices of 5 degree.
thanks!

You can also do it like this:

Let 'x' be the vertices of degree 3

So, (4-x) vertices are of degree >=4 (because 4 vertices have degree >=3)

So,

3(10) + 1(8) + 2(5) + x(3)  + (4-x)(4) >= 62

$\therefore$  2 >= x

Yes, thats why  I have taken >= 62