If ^(n+2)C_8:^(n-2)P_4=57:16, then the value of n/2 is

Correct answer: 9.5. (n+2!)/(8!n-6!)×(n-6!)/(n-2!)=57/16 ⇒n+2n+1nn-1=57/16×8! ⇒n-1nn+1n+2=57×7×6×5×4×3 ⇒n-1nn+1n+2=18×19×20×21 ⇒n=19 ⇒n/2=9.5

Q18 / 98 · NTA Abhyas (JEE Main) 2020 · Mathematics · Permutation Combination

If $^{n + 2} C_{8} :^{n - 2} P_{4} = 57 : 16$ , then the value of $\frac{n}{2}$ is

Correct answer: 9.5

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

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

\Rightarrow (n - 1) n (n + 1) (n + 2) = 57 \times 7 \times 6 \times 5 \times 4 \times 3

\Rightarrow (n - 1) n (n + 1) (n + 2) = 18 \times 19 \times 20 \times 21

\Rightarrow n = 19

\Rightarrow \frac{n}{2} = 9.5

Hence, the correct answer is 9.5.