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Question

C0.C2+C1.C3+C2.C4+........+Cn−2.Cn

A
(2n+1)!(n1)!(n+2)!
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B
(2n)!(n3)!(n+3)!
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C
(2n)!(n2)!(n+2)!
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D
None
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Solution

The correct option is D (2n)!(n2)!(n+2)!
We have
Co+C1x+C2x2+.....+Cnxn=(1+x)n ……..(1)
Also
Coxn+C2xn2+.....+Cn=(x+1)n ……..(2)
Multiplying (1) & (2)
(Co+C1x+C2x2+.....+Cnxn)×(Coxn+C2xn2+....+Cn)=(1+x)2n ………..(3)
Equating coefficients of xnr from both sides of (3), we get
CoCr+C1Cr+1+C2Cr+2+.....+CnrCn
=2nCnr=(2n)!(nr)!(n+2)!
2nCn2=(2n)!(n2)!(n+2)!
Option C is correct.

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