If nC2+nC3=n+1Cr, then the minimum possible value of r is
A
0
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B
3
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C
2
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D
1
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Solution
The correct option is D1 Given nC2+nC3=n+1Cr We know that, nCr+nCr−1=n+1Cr So, nC2+nC3=n+1Cr⇒n+1C3=n+1Cr⇒r=3 or r+3=n+1∴r=3,n−2 From the given expression we know that, n≥3 So, the minimum possible value of r is 1 when n=3.