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Question

A complete graph on n vertices is an undirected graph in which every pair of distinct vertices is connected by an edge. A simple path in a graph is one in which no vertex is repeated. Let $G$ be a complete graph on $10$ vertices. Let $u, v,w$ be three distinct vertices in $G$.How many simple paths are there from $u$ to $v$ going through $w$?

My Approach

I first selected $u,v$ from $10$ vertices by $$\binom{10}{2}$$ then $w$ from remianing $8$ by $$\binom{8}{1}$$

So my possible number of simple path from $u$ to $v$ = $$\binom{10}{2}*\binom{8}{1}*7!$$

But answer seems wrong.

Where am i wrong? Please help me out

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  • $\begingroup$ You dont need $\binom{10}{2}$, $\endgroup$ Sep 22 '17 at 10:48
  • $\begingroup$ @DjuraMarinkov why not $\binom {10}{2}$ , we will not select $u$ and $v$ from avaliable $10$vertex? $\endgroup$
    – laura
    Sep 22 '17 at 10:52
  • $\begingroup$ are not them already defined? Why would you choosing them? You have some specific vertices u,v,w, you dont choose them you choose other vetices. And also I dont see path must go through all vertices $\endgroup$ Sep 22 '17 at 11:31
  • $\begingroup$ I see what you are doing, you are choosing which of them you want to call u,v,w. I dont think it is what is needed. Your path starts wit a specific vertex named u and ends with specific vertex named v. And also it contains vertex named w. That is why I think path may be less than 10 vertices long. Otherwise every path would contain w anyway $\endgroup$ Sep 22 '17 at 11:37
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According to the text we may assume that nodes $u,v,w$ are given, fixed and we have to count the number of simple paths from $u$ to $v$ via $w$.

We consider simple paths according to their length $l$, which is the number of edges of the path.

The path with minimum length contains the nodes of $u,v,w$ only and has length $2$. A path with maximum length contains all $n(=10)$ nodes and has length $n-1=9$. We denote the number of valid paths of length $l$ with $N_l$. The wanted number of all valid paths is \begin{align*} \sum_{l=2}^9N_l \end{align*}

  • Length $l=2$: There is only one path $((u,w),(w,v))$ with length $2$, so $N_2=1$.

  • Length $l=3$: We have $n-3$ ways to select a node $x$. We have two ways to place $w$, either between $u$ and $x$ or between $x$ and $w$. It follows $N_3=2(n-3)$.

  • Length $l=4$: We have $(n-3)(n-4)$ ways to select two nodes and have $3$ ways to place $w$ in between.

Continuing this way we obtain with $n=10$ \begin{align*} &\color{blue}{N_2+N_3+\cdots N_9}\\ &\qquad=1+2(10-3)+3(10-3)(10-4)+4(10-3)(10-4)(10-5)\\ &\qquad\qquad+\cdots+8(10-3)(10-4)\cdots(10-8)\\ &\qquad=\sum_{l=2}^9(l-1)\frac{(10-3)!}{(10-l-1)!}\\ &\qquad\color{blue}{=95\,901} \end{align*} The final result was calculated with some help of Wolfram Alpha.

Note: If we consider these numbers with increasing $n=3,4,\ldots,\color{blue}{10},\ldots$ we obtain \begin{align*} 1,3,11,49,261,1\,631,11\,743,\color{blue}{95\,901},\ldots \end{align*} This sequence is archived as OEIS/A001339.

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  • $\begingroup$ Length $l=2$: There is only one path $((u,v),(v,w))$ with length $2$, so $N_2=1$. Here it should be $((u,w),(w,v))$ ...right? because w is intermediate i guess! $\endgroup$
    – laura
    Sep 22 '17 at 12:00
  • $\begingroup$ @Laura: Thanks! Typo corrected. $\endgroup$
    – epi163sqrt
    Sep 22 '17 at 12:02
  • $\begingroup$ Yes, we both have kinda bumpy explanation. It's simply take k additional vertices and permutate them along with w $\endgroup$ Sep 22 '17 at 20:57
  • $\begingroup$ @DjuraMarinkov: I do not see that great difference. When looking at your first expression, then the right-hand side corresponds to your comment above, while the left-hand side are the two lines of explanation in your answer. I would consider both explanations nearly of the same value. $\endgroup$
    – epi163sqrt
    Sep 23 '17 at 16:45
  • $\begingroup$ It is same, just wanted to point on no need to separately permutate additional vertices and then adding $w$ between. The second explanation is smoother $\endgroup$ Sep 23 '17 at 17:01
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I guess it may not be passing through all the vertices, so simplest path is (u,w,v), another path may be (u,x,w,v)...

$\sum_{k=0}^{7}(k+1)\cdot\binom{7}{k}k!=\sum_{k=0}^{7}\binom{7}{k}(k+1)!$

So, $k$ is number of vertices not named $u,v$ or $w$.

$k+1$ is position of $w$ between the aditional vertices.

$\binom{7}{k}$ is to pick the vertices and $k!$ is to permutate them.

If path has to go through all of them then it is just formula for k=7, $\binom{7}{7}(7+1)!=8!$

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  • $\begingroup$ I just now recognized that our solutions are essentially the same. (+1) $\endgroup$
    – epi163sqrt
    Sep 22 '17 at 16:07

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