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

C0Cr+C1Cr+1+C2Cr+2++CnrCn is equal to 

  1. A

    (2n)!(nr)!(n+r)!

  2. B

    n!r!(n+r)!

  3. C

    n!(nr)!

  4. D

    None of these 

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    Solution:

    We know,

    and (1+x)n=C0+C1x+C2x2++Crxr+(i)1+1xn=C0+C11x+C21x2++Cr1xr+Cr+11xr+1+Cr+21xr+2Cn1xn (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)!(nr)!(n+r)!
     

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