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$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is equal to

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n+1Cr

nCr−1

n−1Cr−1

n+2Cr

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detailed solution

Correct option is D

We have nCr+2nCr−1+nCr−2= nCr+nCr+1+ nCr−j+nCr−2=n+1Cr+n+1Cr−1=n+2Cγ

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nCr+2nCr−1+nCr−2 is equal to

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$^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2}$ is equal to