Pure binomial

$\Large \sum { \sum _{ 0\le i\le j\le 10 }^{ }{ ^{ 10 }{ { C }_{ j }\quad }^{ j }{ { C }_{ i } } } }$

$\Large =\sum _{ j=0 }^{ 10 }{ \sum _{ i=0 }^{ j }{ { { ^{ 10 }{ C } }_{ j }\quad { ^{ j }{ C } }_{ i } } } }$

As ${ ^{ 10 }{ C } }_{ j }$ is independent of variable $i$ we can take it outside the inner sigma.

$\Large =\sum _{ j=0 }^{ 10 }{ { ^{ 10 }{ C } }_{ j }\sum _{ i=0 }^{ j }{ { { ^{ j }{ C } }_{ i } } } }$

Using $\sum _{ i=0 }^{j }{ { ^{ j }{ C } }_{ i }} ={ 2 }^{ j }$ ,

$\Large =\sum _{ j=0 }^{ 10 }{ { ^{ 10 }{ C } }_{ j }{ 2 }^{ j } }$

Using ${ (1+x) }^{ 10 }=\sum _{ r=0 }^{ 10 }{ { ^{ 10 }{ C } }_{ r }{ x }^{ r } }$ ,

$\Large =\sum _{ j=0 }^{ 10 }{ { ^{ 10 }{ C } }_{ j }{ 2 }^{ j } } ={ 3 }^{ 10 }$

$\LARGE =\boxed { {3}^{10}}$

A (10 choose j)×(j choose i) is a coefficient of x^jy^i in the polynomial (x+y+1)^10, hence we get a above sum by taking x=1,y=1. The answer is the last two digit of 3^10

while the understanding of problem statement is the real challenging task,the rest functioning is quite easy.Loop j from 0 to 10,and i loops in a nested form from 0 to j.Now put the combinatorial signs and you are there.