JEE 2023 — Mathematics PYQ

Solution:

1. Formula Use karein: ${}^n{P}_r=\frac{n!}{(n-r)!}$.

   $$\frac{\frac{(2n+1)!}{(2n+1-(n-1))!}}{\frac{(2n-1)!}{(2n-1-n)!}}=\frac{11}{21}$$

2. Simplify Factorials:

   $$\frac{(2n+1)!}{(n+2)!}\cdot \frac{(n-1)!}{(2n-1)!}=\frac{11}{21}$$
   $$\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)n(n-1)!}\cdot \frac{(n-1)!}{(2n-1)!}=\frac{11}{21}$$

3. Terms cancel karein:

   $$\frac{(2n+1)(2n)}{(n+2)(n+1)n}=\frac{11}{21}⟹\frac{2(2n+1)}{(n+2)(n+1)}=\frac{11}{21}$$

4. Equation solve karein:

   $$42(2n+1)=11(n^2+3n+2)⟹84n+42=11n^2+33n+22$$
   $$11n^2-51n-20=0$$

   Factors: $(n-5)(11n+4)=0⟹n=5$ (kyunki $n$ natural number hona chahiye).

5. Final Value:

   $$n^2+n+15=5^2+5+15=25+5+15=45$$

Answer: 45