please explean how to do this ?? Find a combinatorial argument to show that (2n n) = n I = 0 (n i)2. Solution take RHs : (nC0)^2 + (nc1)^2 + (nC3)^2 ..... (ncn)^2 ....::::: now consider expansion of (1 + x)^n = nc0 + nc1*x + nc2*x^2 ...ncn x^n ......; (1 + 1/x)^n = nc0 + nc1/x + nc2/x*x ...ncn .......   multiply both eq :.... (1 + x)^n * (1 + 1/x)^n = (nc0 + nc1*x + nc2*x*x + ..ncn*x^n) * (nc0 + nc1/x + nc2/x*x ...ncn/x^n ) :: in right side when we multiply we also get : terms nc0*nc0 + nc1*nc1 + nc2*nc2 ..... + ncn*ncn . if you see theese are terms devoid of x terms as x/x cancels out .. so try to seek the term which does not have x part in the expansion of left hand side : (1 + x)^n * (1 + 1/x)^n ::= (1 + 1/x + x + 1)^n : the term is 2n C n henc proved .

1. please explean how to do this ??
Find a combinatorial argument to show that (2n n) = n I = 0 (n i)2.
Solution
take RHs : (nC0)^2 + (nc1)^2 + (nC3)^2 ..... (ncn)^2 ....::::: now consider expansion of (1 + x)^n
= nc0 + nc1*x + nc2*x^2 ...ncn x^n ......; (1 + 1/x)^n = nc0 + nc1/x + nc2/x*x ...ncn ....... Â
multiply both eq :.... (1 + x)^n * (1 + 1/x)^n = (nc0 + nc1*x + nc2*x*x + ..ncn*x^n) * (nc0 +
nc1/x + nc2/x*x ...ncn/x^n ) :: in right side when we multiply we also get : terms nc0*nc0 +
nc1*nc1 + nc2*nc2 ..... + ncn*ncn . if you see theese are terms devoid of x terms as x/x cancels
out .. so try to seek the term which does not have x part in the expansion of left hand side : (1 +
x)^n * (1 + 1/x)^n ::= (1 + 1/x + x + 1)^n : the term is 2n C n henc proved