Q.

The value of   nCr−1 nCr(r+1) Cr+1   n+2 nCr nCr+1(r+2) Cr+2   n+2 nCr+1 nCr+2(r+3) Cr+3   n+2  is

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a

n2+n−1

b

0

c

n+3Cr+3

d

nCr−1+nCr+nCr+1

answer is B.

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Detailed Solution

Δ= nCr−1 nCr(r+1)n+2Cr+1 nCr nCr+1(r+2)n+2Cr+2 nCr+1 nCr+2(r+3)n+2Cr+3Applying C1→C1+C2 and using  nCr=nr Cr−1   n-1 in C3, we getΔ= n+1Cr nCr(n+2)n+1Cr n+1Cr+1 nCr+1(n+2)n+1Cr+1 n+1Cr+2 nCr+2(n+2)n+1Cr+2=(n+2) n+1Cr nCr n+1Cr n+1Cr+1 nCr+1 n+1Cr+1 n+1Cr+2 nCr+2 n+1Cr+2=0                           as C1 and C3 are identical)
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