Mixture Again

$\sum _{ 0\le i,j\le n }^{ }{ { C }_{ i }^{ 2 }{ C }_{ j }^{ 2 } }$ $=(\sum _{ i=0 }^{ n }{ { C }_{ i }^{ 2 }) } (\sum _{ j=0 }^{ n }{ { C }_{ j }^{ 2 }) }$

Since $i$ and $j$ are independent of each other

We know, $(\sum _{ i=0 }^{ n }{ { C }_{ i }^{ 2 }) } =\begin{pmatrix} 2n \\ n \end{pmatrix}$

${ \begin{pmatrix} 2n \\ n \end{pmatrix} }^{ 2 }=4900$

${ \begin{pmatrix} 2n \\ n \end{pmatrix} }=70$

$n=4$

Magnitude of $\overset { \rightarrow }{ L }$ will be ${ i }^{ 2 }+{ j }^{ 2 }+k^{ 2 }$

We have to find $\sum _{ 1\le i,j,k\le 4 }^{ }{ { i }^{ 2 }+{ j }^{ 2 }+k^{ 2 } }$

Since all of the variables here independent of each other except for the fact that the number of times one variable's value repeats depend on the other two

If we fix the value of $i$ then $j$ and $k$ can be chosen in $16$ ways, and same goes for fixing $j$ or $k$ ,and since there is symmetry, the sums in all three

cases is equal,so the sum equals, $(1+4+9+16)*(3)*(16)$ which equals $1440$

Required answer is Sum- $1350=1440-1350=90$