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this question is from discrete and combatinatoral mathematics . by grimaldi. i r

ID: 3145599 • Letter: T

Question

this question is from discrete and combatinatoral mathematics . by grimaldi.

i refered chegg solution and that's the contetns....

I can't understand that statements "The total num of paths of length 2 with starting vertex as 'a' can be given as 3 C 2...

why??

there're more than 3C2...

ex..

from vertex 'a'?
a - b - e
a - c - f
a - c- d
a - d - c
a - d - f
isn't it??

why 3C2 ??
I know that like below...
3 C 2 means " select two vertices from (b,c,d) with no ordering"
3C2 = 3 >> bc, bd, cd...

please answer why the answer 3C2 ... T_T..
(I'm not english, so I can't understand english expressions more.. so, please explain carefully... (detailed, precious description please :)
Thank you for reading :)

a) How many different paths of length 2 are there in the undirected graph G in Fig. 11.43? In the above graph, degree of vertex a is 3, that is deg (a) = 3that means there are three edges incident on a and we can choose any of these edges for the path of length two. So the total number of paths of length two with starting vertex as a can be given as Similarly, deg(b)-3; the total number of path of length 2 from b is In the same way we can find the number of paths for remaining vertices Thus, the total number of paths of length two -8-8-8-8-0 =3+3+6+3+1+3 Hence, the total number of paths of length two in the above graphs is 19

Explanation / Answer

Hi,
Let me put the above solution more precisely,
to count the total number of paths in the full graph, lets count all the paths that pass through this vertex,
for example via a:
d-a-c
c-a-b
b-a-c
b-a-d
that is we can choose 2 vertices either side of 'a' so that vertex a passed through them i.e 3C2
now, in the example you gave, if we count paths passing through for all vertices, we can be sure that all vertices will be covered since they one of them can start, pass and end on each one of the path.
hence the answer is 3C2+3C2+4C2+3C2+2C2+3C2=19
Thumbs up if this was helpful, otherwise let me know in comments