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Q.If (n+2)P4=42×nP2(n+2)P_4 = 42 \times {}^nP_2, find nn.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2025Subjective· 3mImportance★★★★★
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Express both permutation terms with factorials; most of the factorial cancels, leaving a quadratic in nn.

(n+2)P4=(n+2)!(n−2)!{}^{(n+2)}P_4=\dfrac{(n+2)!}{(n-2)!} and nP2=n!(n−2)!{}^nP_2=\dfrac{n!}{(n-2)!}.

Given (n+2)P4=42⋅nP2(n+2)P_4=42\cdot{}^nP_2:

(n+2)!(n−2)!=42⋅n!(n−2)!\dfrac{(n+2)!}{(n-2)!}=42\cdot\dfrac{n!}{(n-2)!}

Multiply both sides by (n−2)!(n-2)!: (n+2)!=42⋅n!(n+2)!=42\cdot n!.

Since (n+2)!=(n+2)(n+1)n!(n+2)!=(n+2)(n+1)n!: …

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