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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 ${10 \choose 1}^{2} {10 \choose 2}^{2} {10 \choose 3}^{2} {10 \choose 4}^{2} {10 \choose 5}$.

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.

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  • $\begingroup$ use $\log$ to compute faster. $\endgroup$ – OmG May 31 '17 at 7:14
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Hint $$\sum_{k=0}^{10}\binom{10}{k} = 2^{10}$$

You can proceed with the AM-GM inequality.

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  • $\begingroup$ Very nice idea! $\endgroup$ – Arnab Chatterjee. May 31 '17 at 10:24

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