If ²ⁿ⁺¹ Pₙ₋₁: ²ⁿ⁻¹ Pₙ=11: 21,
then n²+n+15 is equal to :

Solution

Maths sir · 5 steps

Write both as factorials, cancel the big ones against each other, and a neat quadratic in n falls out.

1. Into factorials. Use ᵐPᵣ = m!/(m-r)!. So ²ⁿ⁺¹Pₙ₋₁ = (2n+1)!/(n+2)! and ²ⁿ⁻¹Pₙ = (2n-1)!/(n-1)!.
2. Cancel and simplify. The ratio is (2n+1)!/(2n-1)! × (n-1)!/(n+2)!. The first fraction is (2n+1)(2n) and the second is 1/(n+2)(n+1)n. The n cancels.

   2(2n+1)/(n+1)(n+2) = 11/21

3. Solve the quadratic. Cross-multiply: 42(2n+1) = 11(n²+3n+2), so 11n² - 51n - 20 = 0. The discriminant is 51² + 880 = 3481 = 59², giving n = (51+59)/22 = 5. The other root is negative, so it is rejected.
4. Check. At n = 5 the ratio is (2 × 11)/(6 × 7) = 22/42 = 11/21, as required.
5. So. n² + n + 15 = 25 + 5 + 15 = 45.

The idea to keep. With a ratio of two permutations, write both in factorial form and pair the factorials that are close to each other. Only a few factors survive, and you get a small polynomial equation.

Answer from the NTA final answer key. Solution by PrepWiser.