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