how to find path of length 3 in below graph including V5 ?

Question

what is the issue in the given approach ?

Since the path should pass through v5 I am assuming we only include the path in which v5 can be intermediate vertex.
And path visited in reverse order is not counted as distinct.

Let us consider that vertex v5 is in position 2. Actually it doesn't matter whether vertex v2 is in position 2 or 3 because v2 comes at position 3 in reverse order.

_ v5 _ _

For First vertex we can have four choices. ie v1,v2,v3 and v4

Let us calculate all the possibilities with v1 as first vertex.

We have v1 v5 _ _

For third vertex we can have 3 choices. In this case we can have v2, v3 or v4 as third vertex

When we have v2 as third vertex we have only one choice that is v3.
When we have v3 as third vertex we have 2 choices that is v2 or v4.
When we have v4 as third vertex we have only one choice that is v3

So we have four choices when we start with vertex v1.

For all the four choices for the first vertex we can have total of 16 choices.

This gives total paths possible as 16.

But answer given is 4C3 .

Answer 1

This is a combinatory qs basically. There are 5 vertices ,in which  V5 is fixed,cause it must remain in the path. Now for path of length 3,we need 4 vertices. V5 is already fixed,so we are left with 3 more vertices,which can be chosen from out of 4 vertices.so the  no of paths are 4C3 = 4.

Comments

what is wrong in the given approach ?