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Question

If n+2C6n2P2=11, then n satisfies the equation:

A
n2+2n80=0
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B
n2+5n84=0
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C
n2+3n108=0
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D
n2+n110=0
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Solution

The correct option is C n2+3n108=0
n+2C6n2P2=11n+2C6=11.n2P2(n+2)!6!(n4)!=11.(n2)!(n4)!(n+2)(n+1)(n)(n1)=11×6!(n+2)(n+1)(n)(n1)=11×10×9×8n=9which satisfies n2+3n108=0

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