How many copies of a graph $H$ containing a fixed edge in $[n]$?
Given a graph $H$. For a fixed edge $e$ in a complete graph $K_n$, how many copies of $H$ containing $e$ in $K_n$? Assume $n$ is large enough?
I am wondering if $n^{v_H-2}$ is an upper bound? Since an upper bound I can come up with is $binom{n-2}{v_H-2}v_H!$. But it seems larger than $n^{v_H-2}$ when $v_Hge 3$ and $n$ large enough.
combinatorics graph-theory
asked Nov 19 '18 at 3:17
Connor
865513
add a comment |
Given a graph $H$. For a fixed edge $e$ in a complete graph $K_n$, how many copies of $H$ containing $e$ in $K_n$? Assume $n$ is large enough?
I am wondering if $n^{v_H-2}$ is an upper bound? Since an upper bound I can come up with is $binom{n-2}{v_H-2}v_H!$. But it seems larger than $n^{v_H-2}$ when $v_Hge 3$ and $n$ large enough.
combinatorics graph-theory
asked Nov 19 '18 at 3:17
Connor
865513
Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28
add a comment |
Given a graph $H$. For a fixed edge $e$ in a complete graph $K_n$, how many copies of $H$ containing $e$ in $K_n$? Assume $n$ is large enough?
I am wondering if $n^{v_H-2}$ is an upper bound? Since an upper bound I can come up with is $binom{n-2}{v_H-2}v_H!$. But it seems larger than $n^{v_H-2}$ when $v_Hge 3$ and $n$ large enough.
combinatorics graph-theory
asked Nov 19 '18 at 3:17
Connor
865513
Given a graph $H$. For a fixed edge $e$ in a complete graph $K_n$, how many copies of $H$ containing $e$ in $K_n$? Assume $n$ is large enough?
I am wondering if $n^{v_H-2}$ is an upper bound? Since an upper bound I can come up with is $binom{n-2}{v_H-2}v_H!$. But it seems larger than $n^{v_H-2}$ when $v_Hge 3$ and $n$ large enough.
combinatorics graph-theory
combinatorics graph-theory
asked Nov 19 '18 at 3:17
Connor
865513
asked Nov 19 '18 at 3:17
Connor
865513
asked Nov 19 '18 at 3:17
Connor
865513
asked Nov 19 '18 at 3:17
Connor
865513
asked Nov 19 '18 at 3:17
Connor
865513
865513
Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28
add a comment |
Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28
Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28
Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28
add a comment |
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Isn't it exactly $$binom n{v_H}cdotfrac{v_H!}{|operatorname{Aut}(H)|}cdotfrac{e_H}{binom n2}$$ where $v_H=|V(H)|$ and $e_H=|E(H)|$?
– bof
Nov 19 '18 at 12:28