Why is the denominator of the first term and not ?

The denominator in is . Here and , so .

Can be negative or a fraction?

No, by definition of permutations , and must be non-negative integers with .

JEE MainMathematics

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Permutations and Combinations: Mathematics | JEE Main

{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 21

, then

n^2+n+15

is equal to :

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Step-by-step solutionView

Correct answer

The value of $n^2+n+15$ is 45.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given ratio: ${}^{2n+1}\mathrm{P}_{n-1} : {}^{2n-1}\mathrm{P}_n = 11 : 21$ , where $n \in \mathbb{N}$ .

02goal

Determine the value of $n$ and then compute the target expression $n^2 + n + 15$ .

03approach

Express the permutation terms using the factorial formula ${}^n\mathrm{P}_r = \frac{n!}{(n-r)!}$ , simplify the factorials by canceling common factors, solve the resulting quadratic equation for positive integer $n$ , and substitute into $n^2 + n + 15$ .

04execute

Apply the definition ${}^n\mathrm{P}_r = \frac{n!}{(n-r)!}$ : ${}^{2n+1}\mathrm{P}_{n-1} = \frac{(2n+1)!}{((2n+1)-(n-1))!} = \frac{(2n+1)!}{(n+2)!}$ ${}^{2n-1}\mathrm{P}_n = \frac{(2n-1)!}{((2n-1)-n)!} = \frac{(2n-1)!}{(n-1)!}$ Set up the ratio: $\frac{(2n+1)!}{(n+2)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$

05execute

Expand factorials to cancel identical factors: $\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)n(n-1)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$ $\frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{11}{21}$ Since $n > 0$ , cancel $n$ : $\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$ $\frac{4n+2}{n^2+3n+2} = \frac{11}{21}$

06execute

Cross-multiply and form the quadratic equation: $21(4n + 2) = 11(n^2 + 3n + 2)$ $84n + 42 = 11n^2 + 33n + 22$ $11n^2 - 51n - 20 = 0$ Factor the quadratic equation: $11n^2 - 55n + 4n - 20 = 0$ $11n(n - 5) + 4(n - 5) = 0$ $(11n + 4)(n - 5) = 0$ Since $n$ must be a positive integer, $n = 5$ .

07execute

Evaluate $n^2 + n + 15$ at $n = 5$ : $n^2 + n + 15 = 5^2 + 5 + 15 = 25 + 5 + 15 = 45$ .

✓verify

Substitute $n=5$ into the original ratio: ${}^{11}\mathrm{P}_4 / {}^{9}\mathrm{P}_5 = \frac{11 \times 10 \times 9 \times 8}{9 \times 8 \times 7 \times 6 \times 5} = \frac{7920}{15120} = \frac{11}{21}$ . The ratio matches identically.

Hints that build this answer step by step

How is the permutation formula ${^n\mathrm{P}_r} = \frac{n!}{(n-r)!}$ applied to expand the ratio $\frac{{^{2n+1}\mathrm{P}_{n-1}}}{{^{2n-1}\mathrm{P}_n}}$ ?
$\frac{(2n+1)!}{(n+2)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$
After expanding $(2n+1)! = (2n+1)(2n)(2n-1)!$ and $(n+2)! = (n+2)(n+1)n(n-1)!$ , what simplified equation in $n$ is obtained?
$\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$
Cross-multiplying $\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$ yields which quadratic equation?
$11n^2 - 51n - 20 = 0$
Factoring $11n^2 - 51n - 20 = (11n+4)(n-5) = 0$ gives $n=5$ . What is the final value of $n^2 + n + 15$ ?
$45$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the denominator of the first term

(n+2)!

and not

n!

The denominator in ${}^n\mathrm{P}_r$ is $(n-r)!$ . Here $n = 2n+1$ and $r = n-1$ , so $(2n+1) - (n-1) = 2n + 1 - n + 1 = n + 2$ .

Can

n

be negative or a fraction?

No, by definition of permutations ${}^n\mathrm{P}_r$ , $n$ and $r$ must be non-negative integers with $n \ge r$ .

Answer

The value of $n^2+n+15$ is 45.

Why each option works or fails

Step-by-step solution

given: Given ratio: ${}^{2n+1}\mathrm{P}_{n-1} : {}^{2n-1}\mathrm{P}_n = 11 : 21$ , where $n \in \mathbb{N}$ .
goal: Determine the value of $n$ and then compute the target expression $n^2 + n + 15$ .
approach: Express the permutation terms using the factorial formula ${}^n\mathrm{P}_r = \frac{n!}{(n-r)!}$ , simplify the factorials by canceling common factors, solve the resulting quadratic equation for positive integer $n$ , and substitute into $n^2 + n + 15$ .
execute: Apply the definition ${}^n\mathrm{P}_r = \frac{n!}{(n-r)!}$ : ${}^{2n+1}\mathrm{P}_{n-1} = \frac{(2n+1)!}{((2n+1)-(n-1))!} = \frac{(2n+1)!}{(n+2)!}$ ${}^{2n-1}\mathrm{P}_n = \frac{(2n-1)!}{((2n-1)-n)!} = \frac{(2n-1)!}{(n-1)!}$ Set up the ratio: $\frac{(2n+1)!}{(n+2)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$
execute: Expand factorials to cancel identical factors: $\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)n(n-1)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$ $\frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{11}{21}$ Since $n > 0$ , cancel $n$ : $\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$ $\frac{4n+2}{n^2+3n+2} = \frac{11}{21}$
execute: Cross-multiply and form the quadratic equation: $21(4n + 2) = 11(n^2 + 3n + 2)$ $84n + 42 = 11n^2 + 33n + 22$ $11n^2 - 51n - 20 = 0$ Factor the quadratic equation: $11n^2 - 55n + 4n - 20 = 0$ $11n(n - 5) + 4(n - 5) = 0$ $(11n + 4)(n - 5) = 0$ Since $n$ must be a positive integer, $n = 5$ .
execute: Evaluate $n^2 + n + 15$ at $n = 5$ : $n^2 + n + 15 = 5^2 + 5 + 15 = 25 + 5 + 15 = 45$ .
verify: Substitute $n=5$ into the original ratio: ${}^{11}\mathrm{P}_4 / {}^{9}\mathrm{P}_5 = \frac{11 \times 10 \times 9 \times 8}{9 \times 8 \times 7 \times 6 \times 5} = \frac{7920}{15120} = \frac{11}{21}$ . The ratio matches identically.

Permutations and Combinations: Mathematics | JEE Main

What feels right?

The value of $n^2+n+15$ is 45.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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