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Question

Let f(n)=∣ ∣ ∣nn+1n+2nPnn+1Pn+1n+1Pn+2nCnn+1Cn+1n+1Cn+2∣ ∣ ∣
Then, f(n) is divisible by


A

n2+n+1

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B

(n+1)!

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C

n!

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D

none of these

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Solution

The correct options are
A

n2+n+1


C

n!


f(n)=∣ ∣ ∣nn+1n+2nPnn+1Pn+1n+2Pn+2nCnn+1Cn+1n+2Cn+2∣ ∣ ∣=n!∣ ∣ ∣n(n+1)(n+2)1(n+1)(n+1)(n+2)111∣ ∣ ∣Δ=n!∣ ∣ ∣n111n(n+1)2100∣ ∣ ∣(C3C3C2,C2C2C1)Δ=n!×[(n+1)2n]=n!×(n2+n+1)


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