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The question is to find which of the following option is correct regarding $$\left(\frac{\ \ 2^{10}}{11}\right)^{11}$$

$A)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}$

$B)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2$ but strictly smaller than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2 \binom{10}{5}$

$C)$ less than or equal to $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2$

$D)$ equal to $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}$



This question came in I.S.I. B.Stat 2013 Entrance exam in which calculators were not allowed.I tried using some combaratorial argument but failed.Any help shall be highly appreciated.Thanks.

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Is this AM-GM? We have $$\binom{10}1^2\binom{10}2^2\binom{10}3^2\binom{10}4^2\binom{10}5 =\prod_{k=0}^{10}\binom{10}k<\left(\frac1{11}\sum_{k=0}^{10} \binom{10}k\right)^{11}=\left(\frac{2^{10}}{11}\right)^{11}. $$

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  • $\begingroup$ Wow!. BTW, a typo, k = 0 to 10 in the product. $\endgroup$ – user94300 Apr 28 '17 at 6:47

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