Q.

The value of  nC0×2nCrnC1×2n2Cr+nC2×2n4Cr+ is equal to

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a

 nCrn×2nr if rn

b

0, if r<n

c

nCrn×22nr if r<n

d

 nCrn×22nr if rn

answer is A, B.

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Detailed Solution

 nC0×2nCrnC1×2n2Cr+nC2×2n4Cr+

              = Coefficient of xr in  nC0(1+x)2nnC1(1+x)2n2+nC2(1+x)2n4+ = Coefficient of xr in  nC0(1+x)2nnC1(1+x)2n1+nC2(1+x)2n2+ = Coefficient of xr in (1+x)21n = Coefficient of xr in 2x+x2n

General term, Tp+1=nCp(2x)npx2p=nCp2npxn+p

For xr,n+p=r. So, p=rn

 Coefficient of xr=nCrn×22nr

This is non-zero if rn and zero if r < n.

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