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Question

Question: The value of <sup>n</sup>C<sub>0</sub> .<sup>2n</sup>C<sub>r</sub> – <sup>n</sup>C<sub>1</sub> .<su...

The value of

nC0 .2nCrnC1 .2n – 2Cr + nC2 .2n – 4Cr +......... + (–1)nnCn2n – 2nCn is –

A

{22nrnCrn,ifr>n0,ifr<n \left\{ \begin{matrix} {2^{2n - r}}^{n}C_{r - n} & , & ifr > n \\ 0 & , & ifr < n \end{matrix} \right.\

B

{32nrnCrn,ifr>n0,ifr<n \left\{ \begin{matrix} 3^{2n - rn}C_{r - n} & , & ifr > n \\ 0 & , & ifr < n \end{matrix} \right.\

C

{22nrnCr+n,ifr>n0,ifr<n \left\{ \begin{matrix} {2^{2n - r}}^{n}C_{r + n} & , & ifr > n \\ 0 & , & ifr < n \end{matrix} \right.\

D

None of these

Answer

{22nrnCrn,ifr>n0,ifr<n \left\{ \begin{matrix} {2^{2n - r}}^{n}C_{r - n} & , & ifr > n \\ 0 & , & ifr < n \end{matrix} \right.\

Explanation

Solution

We have

nC0 . 2nCrnC1 . 2n–2Cr + nC2 . 2n–4Cr + ....+ (–1)n nCn 2n–2nCn

= Coefficient of xr in [nC0 (1 + x)2nnC1 (1 + x)2n–2 + nC2

(1 + x)2n–4 – ….…+ (–1)n nCn (1 + x)2n–2n]

= Coefficient of xr in [(1 + x)2 – 1]n

= Coefficient of xr in (2x + x2)n

= Coefficient of xr–n in (2 + x)n

= {22nrnCrn,ifr>n0,ifr<n \left\{ \begin{matrix} 2^{2n - rn}C_{r - n} & , & ifr > n \\ 0 & , & ifr < n \end{matrix} \right.\

Hence (1) is correct answer.