What is the upper bound of this quantity given $p_1+p_2+p_3=n$? (or $2p_1+p_2=n, p_3=0$ or $3p_1=n, p_2=p_3=0$)?

$$\frac{n!}{p_1!p_2!p_3!}\cdot \exp\left(-\frac{\binom{n}{2}-\binom{p_1}{2}-\binom{p_2}{2}-\binom{p_3}{2}}{2\beta(n)}\right)$$ where $n$ is positive integer, and $\beta(n)$ is a increasing function of $n$.

What I tried is to use Stirling formula to approximate the quantity thus obtained $$\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\sqrt{2\pi p_1}\left(\frac{p_1}{e}\right)^{p_1} \cdot \sqrt{2\pi p_2}\left(\frac{p_2}{e}\right)^{p_2} \cdot \sqrt{2\pi p_3}\left(\frac{p_3}{e}\right)^{p_3}} \cdot \exp\left(-\frac{\binom{n}{2}-\binom{p_1}{2}-\binom{p_2}{2}-\binom{p_3}{2}}{2\beta(n)}\right)$$

However I have no idea of how to proceed...

Hope anyone could help me out. Thanks in advance for any comment!



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