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Question

# If nCr+4nCr+1+6nCr+2+4nCr+3+nCr+4nCr+3nCr+1+3nCr+2+nCr+3=n+kr+k, then the value of k is

A
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B
2
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C
4
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D
5
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Solution

## The correct option is C 4Use nCr+nCr−1=n+1Cr repeatedly.NumeratornCr+4nCr+1+6nCr+2+4nCr+3+nCr+4=(nCr+nCr+1)+3(nCr+1+nCr+2)+3(nCr+2+nCr+3)+(nCr+3+nCr+4)=(n+1Cr+1)+3(n+1Cr+2)+3(n+1Cr+3)+(n+1Cr+4)=(n+1Cr+1+n+1Cr+2)+2(n+1Cr+2+n+1Cr+3)+(n+1Cr+3+n+1Cr+4)=(n+2Cr+2)+2(n+2Cr+3)+(n+2Cr+4)=(n+2Cr+2+n+2Cr+3)+(n+2Cr+3+n+2Cr+4)=n+3Cr+3+n+3Cr+4=n+4Cr+4Similarly, DenominatornCr+3nCr+1+3nCr+2+nCr+3=n+3Cr+3∴nCr+4nCr+1+6nCr+2+4nCr+3+nCr+4nCr+3nCr+1+3nCr+2+nCr+3=n+4Cr+4n+3Cr+3=(n+4)!(r+4)!(n+4−r−4)!(n+3)!(r+3)!(n+3−r−3)!=n+4r+4=n+kr+k∴k=4

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