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I think the following equality is correct, but I'm not sure, so I'm asking you: $$\left(\prod_{\large\tfrac{n}{2}\,<\,p\,\le\,\tfrac{6n}{7}}p\right)\cdot\left(\prod_{\large n\,<\,p\,\le\,3n\strut}p\right)<\binom{3n}{\frac{n}{2}}$$ for even $ n $. $ p $ is a prime.

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Not correct.

For $n=4$, $$1155=3\cdot (5\times 7\times 11)\color{red}{\gt} \binom{12}{2}=66.$$

mathlove
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