# HELP NEEDED !!

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 $\frac { { 2 }^{ 40 }-\left( \begin{matrix} 40 \\ 20 \end{matrix} \right) }{ 2 }$

Note by Vighnesh Raut
5 years, 2 months ago

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

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

$2^{40}=\binom{40}{20}+2\lambda$

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

Note :

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

- 5 years, 2 months ago

Thanks Sir for the detailed solution.

- 5 years, 2 months ago

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

- 5 years, 2 months ago

From where did you get this question?

- 5 years, 2 months ago

It came in my mock mains test..

- 5 years, 2 months ago

Ok,which coaching centre?

- 5 years, 2 months ago

Career Launcher..

- 5 years, 2 months ago

ok thanx!

- 5 years, 2 months ago

where have you joined??

- 5 years, 2 months ago

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

- 5 years, 2 months ago

oh ok....

- 5 years, 2 months ago