COMEDK 2025 Afternoon Solved Paper

Concept:The given ratio of combination and permutation is simplified into an equation involving a product of four consecutive integers.Explanation:Write the terms in factorial form:${}^{n+2}C_8 = \frac{(n+2)!}{8!\,(n-6)!}$ and ${}^{n-2}P_4 = \frac{(n-2)!}{(n-6)!}$.The ratio becomes $\frac{\frac{(n+2)!}{8!\,(n-6)!}}{\frac{(n-2)!}{(n-6)!}} = \frac{(n+2)!}{8!\,(n-2)!}$.Simplify the factorial ratio: $\frac{(n+2)!}{(n-2)!} = (n+2)(n+1)n(n-1)$.So $\frac{(n+2)(n+1)n(n-1)}{8!} = \frac{57}{16}$.Multiply both sides by $8! = 40320$:$(n-1)n(n+1)(n+2) = \frac{40320 \times 57}{16} = 143640$.Now check the options by computing the product of four consecutive numbers starting from $n-1$ to $n+2$:• $n = 17$: $16 \times 17 \times 18 \times 19 = 93024$ (no)• $n = 19$: $18 \times 19 \times 20 \times 21 = 143640$ (yes)• $n = 20$: $19 \times 20 \times 21 \times 22 = 175560$ (no)• $n = 21$: $20 \times 21 \times 22 \times 23 = 212520$ (no)Only $n=19$ satisfies the equation.Answer:$n = 19$ (Option B).