If x + y = 1, then value of ∑r=0n (r) nCrxn−ryr is 

If x + y = 1, then value of r=0n(r) nCrxnryr is 

  1. A

    1

  2. B

    0

  3. C

    nx

  4. D

    ny

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    Solution:

    We have (a+t)n=r=0nnCranrtr 

    Differentiating both the sides with respect to t, we get 

    n(a+t)n1=r=0nr nCranrtr1

    Multiplying both the sides by t and putting a = x and t = y, we get

    n(x+y)n1y=r=0nr nCrxnryr 

    r=0nr nCrxnryr=ny [ x+y=1]

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