TITLE: Choose an interval where the number $\left(\frac {2^{10}} {11}\right)^{11}$ is situated.
QUESTION [6 upvotes]: The number
$$\left(\frac {2^{10}} {11}\right)^{11}$$
is
$(A)$ strictly larger than  ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$.
$(B)$ strictly larger than ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2}$ but strictly smaller than ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$.
$(C)$ less than or equal to ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2}$.
$(D)$ equal to ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$.
Source

My attempt $:$
Since $2^4 > 11$ so we have $\left(\frac {2^{10}} {11}\right)> 2^{6}$ $\implies$ $\left(\frac {2^{10}} {11}\right)^{11}>(2^{6})^{11}$.
Now $(2^{6})^{11}=(2^{6})^{10}.2^{6}=(2^{10})^{6}.2^{6}=(2^{10})^{4}.2^{26}>{10 \choose 3}^{2} {10 \choose 4}^{2}.2^{26}$.Now we will show that $2^{26}>{10 \choose 1}^{2} {10 \choose 2}^{2}{10 \choose 5}$.
Now $2^{7}>100={10 \choose 1}^{2},2^{11}>2025={10 \choose 2}^{2}$ and $2^{8}>252={10 \choose 5}$.This shows that $2^{26}>{10 \choose 1}^{2} {10 \choose 2}^{2}{10 \choose 5}$. i.e. we have $(2^{6})^{11}>{10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$ and consequently $\left(\frac {2^{10}} {11}\right)^{11}>{10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$.Hence $(A)$ is the correct option.
Now my question is :
Is there alternative easy way to determine the above option?Because at last the calculation becomes too much laborious and boring.Please help me.
Thank you in advance.

REPLY [7 votes]: Hint $$\sum_{k=0}^{10}\binom{10}{k} = 2^{10}$$
You can proceed with the AM-GM inequality.