JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Ratio of Permutations and Evaluating a Quadratic Expression

Q: Why is the denominator of the first term (n+2)! and not n!?

The denominator in ^nP_r is (n-r)!. Here n = 2n+1 and r = n-1, so (2n+1) - (n-1) = 2n + 1 - n + 1 = n + 2.

Q: Can n be negative or a fraction?

No, by definition of permutations ^nP_r, n and r must be non-negative integers with n ≥ r.

If ^(2 n+1) P_(n-1): ^(2 n-1) 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: $^{2 n + 1} P_{n - 1} :^{2 n - 1} P_{n} = 11 : 21$ , where $n \in N$ .

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 $^{n} P_{r} = \frac{n!}{(n - r)!}$ , 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 $^{n} P_{r} = \frac{n!}{(n - r)!}$ : $^{2 n + 1} P_{n - 1} = \frac{(2 n + 1)!}{((2 n + 1) - (n - 1))!} = \frac{(2 n + 1)!}{(n + 2)!}$ $^{2 n - 1} P_{n} = \frac{(2 n - 1)!}{((2 n - 1) - n)!} = \frac{(2 n - 1)!}{(n - 1)!}$ Set up the ratio: $\frac{(2 n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

05Execute

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

06Execute

Cross-multiply and form the quadratic equation: $21 (4 n + 2) = 11 (n^{2} + 3 n + 2)$ $84 n + 42 = 11 n^{2} + 33 n + 22$ $11 n^{2} - 51 n - 20 = 0$ Factor the quadratic equation: $11 n^{2} - 55 n + 4 n - 20 = 0$ $11 n (n - 5) + 4 (n - 5) = 0$ $(11 n + 4) (n - 5) = 0$ 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: $^{11} P_{4} /^{9} P_{5} = \frac{11 \times 10 \times 9 \times 8}{9 \times 8 \times 7 \times 6 \times 5} = \frac{7920}{15120} = \frac{11}{21}$ . The ratio matches identically.

Hints that build this answer step by step

How is the permutation formula $^{n} P_{r} = \frac{n!}{(n - r)!}$ applied to expand the ratio $\frac{^{2 n + 1} P_{n - 1}}{^{2 n - 1} P_{n}}$ ?
$\frac{(2 n + 1)!}{(n + 2)!} \cdot \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$
After expanding $(2 n + 1)! = (2 n + 1) (2 n) (2 n - 1)!$ and $(n + 2)! = (n + 2) (n + 1) n (n - 1)!$ , what simplified equation in $n$ is obtained?
$\frac{2 (2 n + 1)}{(n + 2) (n + 1)} = \frac{11}{21}$
Cross-multiplying $\frac{2 (2 n + 1)}{(n + 2) (n + 1)} = \frac{11}{21}$ yields which quadratic equation?
$11 n^{2} - 51 n - 20 = 0$
Factoring $11 n^{2} - 51 n - 20 = (11 n + 4) (n - 5) = 0$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $^{n} P_{r}$ is $(n - r)!$ . Here $n = 2 n + 1$ and $r = n - 1$ , so $(2 n + 1) - (n - 1) = 2 n + 1 - n + 1 = n + 2$ .

Can

n

be negative or a fraction?

No, by definition of permutations $^{n} P_{r}$ , $n$ and $r$ must be non-negative integers with $n \geq r$ .

Answer

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

Why each option works or fails

Step-by-step solution

Given: Given ratio: $^{2 n + 1} P_{n - 1} :^{2 n - 1} P_{n} = 11 : 21$ , where $n \in N$ .
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 $^{n} P_{r} = \frac{n!}{(n - r)!}$ , 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 $^{n} P_{r} = \frac{n!}{(n - r)!}$ : $^{2 n + 1} P_{n - 1} = \frac{(2 n + 1)!}{((2 n + 1) - (n - 1))!} = \frac{(2 n + 1)!}{(n + 2)!}$ $^{2 n - 1} P_{n} = \frac{(2 n - 1)!}{((2 n - 1) - n)!} = \frac{(2 n - 1)!}{(n - 1)!}$ Set up the ratio: $\frac{(2 n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$
Execute: Expand factorials to cancel identical factors: $\frac{(2 n + 1) (2 n) (2 n - 1)!}{(n + 2) (n + 1) n (n - 1)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$ $\frac{(2 n + 1) (2 n)}{(n + 2) (n + 1) n} = \frac{11}{21}$ Since $n > 0$ , cancel $n$ : $\frac{2 (2 n + 1)}{(n + 2) (n + 1)} = \frac{11}{21}$ $\frac{4 n + 2}{n^{2} + 3 n + 2} = \frac{11}{21}$
Execute: Cross-multiply and form the quadratic equation: $21 (4 n + 2) = 11 (n^{2} + 3 n + 2)$ $84 n + 42 = 11 n^{2} + 33 n + 22$ $11 n^{2} - 51 n - 20 = 0$ Factor the quadratic equation: $11 n^{2} - 55 n + 4 n - 20 = 0$ $11 n (n - 5) + 4 (n - 5) = 0$ $(11 n + 4) (n - 5) = 0$ 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: $^{11} P_{4} /^{9} P_{5} = \frac{11 \times 10 \times 9 \times 8}{9 \times 8 \times 7 \times 6 \times 5} = \frac{7920}{15120} = \frac{11}{21}$ . The ratio matches identically.

Ratio of Permutations and Evaluating a Quadratic Expression

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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