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Question

Question: \[nC_{r} + 2^{n} ⥂ C_{r - 1} +^{n} ⥂ C_{r - 2} =\]...

nCr+2n⥂Cr−1+n⥂Cr−2=nC_{r} + 2^{n} ⥂ C_{r - 1} +^{n} ⥂ C_{r - 2} =

A

n+1Crn + 1C_{r}

B

n+1Cr+1n + 1C_{r + 1}

C

n+2Crn + 2C_{r}

D

n+2Cr+1n + 2C_{r + 1}

Answer

n+2Crn + 2C_{r}

Explanation

Solution

nCr+2⥂nCr−1+nCr−2=nCr+nCr−1+nCr−1+nCr−2nC_{r} + 2 ⥂^{n}C_{r - 1} +^{n}C_{r - 2} =^{n}C_{r} +^{n}C_{r - 1} +^{n}C_{r - 1} +^{n}C_{r - 2}

=n+1Cr+n+1Cr−1=n+2Cr=^{n + 1}C_{r} +^{n + 1}C_{r - 1} =^{n + 2}C_{r}.