C0Cr+C1Cr+1+C2Cr+2+…+Cn−rCn is equal to

We know,and (1+x)n=C0+C1x+C2x2+…+Crxr+……(i)1+1xn=C0+C11x+C21x2+…+Cr1xr+Cr+11xr+1+Cr+21xr+2…Cn1xn… (ii) on multiplying Eqs. (i) and (ii), equating coefficient of xr in 1xn(1+x)2n orthecoefficientof xn+r in (1+x)2n, we get the value of required expression which is  2nCn+r=(2n)!(n−r)!(n+r)!