Can anyone please help me in solving this \[\displaystyle{\sum _{ i=0 }^{ 20 }{ \sum _{ j=i+1 }^{ 20 }{ { { \left( \begin{matrix} 20 \\ i \end{matrix} \right) } } } } \left( \begin{matrix} 20 \\ j \end{matrix} \right) }\]

Its answer is \[\frac { { 2 }^{ 40 }-\left( \begin{matrix} 40 \\ 20 \end{matrix} \right) }{ 2 } \]

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TopNewest\[\displaystyle{\sum _{ i=0 }^{ 20 }{ \sum _{ j=i+1 }^{ 20 }{ { { \left( \begin{matrix} 20 \\ i \end{matrix} \right) } } } } \left( \begin{matrix} 20 \\ j \end{matrix} \right) }\] \[= \displaystyle \sum_{i=0}^{20} \binom{20}{i} \times \left[ \binom{20}{i+1}+\binom{20}{i+2}+.....+\binom{20}{20}\right]\] \[=\text{ Sum of the product of every possible combinations of two out of} \binom{20}{0} , \binom{20}{1}, ...., \binom{20}{20}=\lambda (say)\]

\[\left[ \binom{20}{0} +\binom{20}{1}+ ....+ \binom{20}{20}\right]^2= \binom{20}{0}^2+ \binom{20}{1}^2+ ....+ \binom{20}{20}^2 +2 \lambda\]

\[2^{40}=\binom{40}{20}+2\lambda\]

\[\implies \lambda=\dfrac{2^{40}-\binom{40}{20}}{2}\]

Note :

\[\displaystyle \sum_{k=0}^n \binom{n}{k}^2=\binom{2n}{n}\]

@Vighnesh Raut – Sandeep Bhardwaj · 1 year, 11 months ago

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– Harshvardhan Mehta · 1 year, 11 months ago

Thanks Sir for the detailed solution.Log in to reply

– Vighnesh Raut · 1 year, 11 months ago

Thank you so much sir.... It is a very detailed solution...Understood the process..Once again thanks..Log in to reply

From where did you get this question? – Adarsh Kumar · 1 year, 11 months ago

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– Vighnesh Raut · 1 year, 11 months ago

It came in my mock mains test..Log in to reply

– Adarsh Kumar · 1 year, 11 months ago

Ok,which coaching centre?Log in to reply

– Vighnesh Raut · 1 year, 11 months ago

Career Launcher..Log in to reply

– Adarsh Kumar · 1 year, 11 months ago

ok thanx!Log in to reply

– Vighnesh Raut · 1 year, 11 months ago

where have you joined??Log in to reply

– Adarsh Kumar · 1 year, 11 months ago

I am 14 right now(going to be 15 on may 12),I don't go to any coaching centre.Sorry.Log in to reply

– Vighnesh Raut · 1 year, 11 months ago

oh ok....Log in to reply