I was wondering if there was a intuitive/heuristic argument to understand why generalizing the QCD gauge group $SU(3)$ to $SU(N)$ and taking $N\rightarrow \infty$ simplifies the analysis of the theory. Since in this limit only planar diagrams survive, the others being suppressed.

At first I would expect things to get a lot nastier by introducing such large number $N$ of colors...

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    $\begingroup$ It's an expansion parameter, like any other. In the large-N limit, many diagrams become sub-dominant. So it's enough to deal with a dominant diagrams, which are easier to deal with. $\endgroup$ – Siva Mar 11 '15 at 16:17

The intuitive idea is based on the Central Limit Theorem. Because suppose (as is usually the case) that your matter fields are in the fundamental representation of $SU(N)$, then the multiplet contains $N$ independent fields. Now the central limit tells us that the arithmetic average of $N$ independent random variables self-averages to a normal distribution, i.e. have small fluctuations. So in QCD for example, hadrons are always color singlets; a pion is $\pi = \sum_{c = 1}^N q_c\bar{q}_c$, so they have to be an average over all quark colors, so for $N\rightarrow \infty$ the fluctuations of $\pi$ (and hadrons in general) are much smaller than those of $q$ (quarks), because they should self-average according to the CLT. And in particular $$\left\langle \pi(x) \pi(y) \right\rangle \xrightarrow{N\rightarrow\infty}\left\langle \pi(x)\right\rangle\left\langle \pi(y)\right\rangle$$


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