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Question Number 186285 by Shrinava last updated on 03/Feb/23

Prove that: R (m , n) ≤ C_(m+n) ^m Here R states the Ramsey theory

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathrm{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$\mathrm{Here}\:\:\mathrm{R}\:\:\mathrm{states}\:\mathrm{the}\:\:\mathrm{Ramsey}\:\:\mathrm{theory} \\ $$

Commented by mr W last updated on 03/Feb/23

R(m,n)≤R(m−1,n)+R(m,n−1) R(m,n)≤ (((m+n−2)),((m−1)) ) = (((n+m−1)),(m) )− (((n+m−2)),(m) ) = (((n+m)),(m) )− (((n+m−1)),((m−1)) )− (((n+m−2)),(m) ) ≤ (((n+m)),(m) )

$${R}\left({m},{n}\right)\leqslant{R}\left({m}−\mathrm{1},{n}\right)+{R}\left({m},{n}−\mathrm{1}\right) \\ $$$${R}\left({m},{n}\right)\leqslant\begin{pmatrix}{{m}+{n}−\mathrm{2}}\\{{m}−\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\begin{pmatrix}{{n}+{m}−\mathrm{1}}\\{{m}}\end{pmatrix}−\begin{pmatrix}{{n}+{m}−\mathrm{2}}\\{{m}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\begin{pmatrix}{{n}+{m}}\\{{m}}\end{pmatrix}−\begin{pmatrix}{{n}+{m}−\mathrm{1}}\\{{m}−\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{{n}+{m}−\mathrm{2}}\\{{m}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\begin{pmatrix}{{n}+{m}}\\{{m}}\end{pmatrix} \\ $$

Commented by Shrinava last updated on 03/Feb/23

perfect dear professor thank you so much

$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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