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Find n if 16. n+2C8 = 57 n-2P4

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Given,

\(\frac{^{n+2}C_8}{^{n-2}P_4}=\frac{57}{16}\)

\(\Rightarrow \frac{\frac{(n+2)!}{8!(n+2-8)!}}{\frac{(n-2)!}{(n-2-4)!}}=\frac{57}{16}\)

\(\Rightarrow \frac{\frac{(n+2)!}{8!(n+6)!}}{\frac{(n-2)!}{(n-6)!}}=\frac{57}{16}\)

\(\Rightarrow \frac{\frac{(n+2)!}{8!}}{\frac{(n-2)!}{1!}}=\frac{57}{16}\)

\(\Rightarrow \frac{(n+2)!}{(n-2)!}=\frac{57}{16} \times 8!\)

\(\Rightarrow \frac {(n+2)(n+1)(n)(n-1)(n-2)}{(n-2)!}\)

\(\Rightarrow \frac{57}{16}\times8\times7\times6\times5\times4\times3\times2\times1\)

⇒ (n + 2)(n + 1)(n)(n − 1) = 57 × 7 × 6 × 5 × 4 × 3

As L.H.S is the product of four consecutive natural numbers, hence

⇒ (n + 2)(n + 1)(n)(n − 1) = 19 × 3 × 7 × 6 × 5 × 4 × 3

⇒ (n + 2)(n + 1)(n)(n − 1) = 19 × (3 × 7) × (4 × 5) × (6 × 3)

⇒ (n + 2)(n + 1)(n)(n − 1) = 19 × 21 × 20 × 18

On rearranging, we get

⇒ (n + 2)(n + 1)(n)(n − 1) = 19 × 21 × 20 × 18

⇒ n + 2 = 21

⇒ n = 21 − 2 = 19

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