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.


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.