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Let $n$ be divisible by $3$. Prove that

$$c(n, n/3) \geq \frac{n!}{3^{n/3}(n/3)!}$$


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Recall that $c(n, n/3)$ counts the permutations of $\{1, 2, \dots, n\}$ into $n/3$ cycles, while $\frac{n!}{3^{n/3}(n/3)!}$ counts the permutations of $\{1, 2, \dots, n\}$ into $n/3$ 3-cycles.

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