If nP2 = 20 × nP0 , then the value of n is given by:

Option a: n=5. Solution: Let us solve the equation as printed in the question: nP2 = 20 × nP0 Substituting nP2 = n(n-1) and nP0 = 1 into the equation: n(n-1) = 20 × 1 n2 - n - 20 = 0 (n-5)(n+4) = 0 Since n must be a positive integer, we discard n = -4 , leaving: n = 5 Mathematically, the correct answer is n = 5 , which corresponds to Option A. However, the textbook key lists Option C ( n=9 ) as correct. If the textbook key is correct, the equation must have had a typographical error where 72 was replaced by 20: nP2 = 72 × nP0 n(n-1) = 72 n2 - n - 72 = 0 (n-9)(n+8) = 0 n = 9 This matches Option C ( n=9 ). We proceed with Option A as the mathematically correct answer for the printed equation. Hence, Option A is the correct answer.

If $^nP_2 = 20 \cdot ^nP_0$, then the value of $n$ is given by:

The correct answer is Option a: $n=5$. Let us solve the equation as printed in the question: $$^nP_2 = 20 \\cdot ^nP_0$$ Substituting $^nP_2 = n(n-1)$ and $^nP_0 = 1$ into the equation: $$n(n-1) = 20 \\times 1$$ $$n^2 - n - 20 = 0$$ $$(n-5)(n+4) = 0$$ Since $n$ must be a positive integer, we discard $n = -4$, leaving: $$n = 5$$ Mathematically, the correct answer is $n = 5$, which corresponds to Option A. However, the textbook key lists Option C ($n=9$) as correct. If the textbook key is correct, the equation must have had a typographical error where 72 was replaced by 20: $$^nP_2 = 72 \\cdot ^nP_0$$ $$n(n-1) = 72 \\implies n^2 - n - 72 = 0 \\implies (n-9)(n+8) = 0 \\implies n = 9$$ This matches Option C ($n=9$). We proceed with Option A as the mathematically correct answer for the printed equation. Hence, **Option A** is the correct answer.

Permutations and CombinationsMCQPYQ July 21Question 1607 of 251

All Questions

If $\displaystyle ^nP_2 = 20 \cdot ^nP_0$ , then the value of $\displaystyle n$ is given by:

Options

\displaystyle n=5

\displaystyle n=3

\displaystyle n=9

\displaystyle n=8

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle n=5$

All Options:

A $\displaystyle n=5$
B $\displaystyle n=3$
C $\displaystyle n=9$
D $\displaystyle n=8$

Detailed Solution & Explanation

Let us solve the equation as printed in the question:

^nP_2 = 20 \\cdot ^nP_0

Substituting

\displaystyle ^nP_2 = n(n-1)

and

\displaystyle ^nP_0 = 1

into the equation:

n(n-1) = 20 \\times 1

n^2 - n - 20 = 0

(n-5)(n+4) = 0

Since

\displaystyle n

must be a positive integer, we discard

\displaystyle n = -4

, leaving:

n = 5

Mathematically, the correct answer is

\displaystyle n = 5

, which corresponds to Option A. However, the textbook key lists Option C (

\displaystyle n=9

) as correct.

If the textbook key is correct, the equation must have had a typographical error where 72 was replaced by 20:

^nP_2 = 72 \\cdot ^nP_0

n(n-1) = 72 \\implies n^2 - n - 72 = 0 \\implies (n-9)(n+8) = 0 \\implies n = 9

This matches Option C (

\displaystyle n=9

). We proceed with Option A as the mathematically correct answer for the printed equation.
Hence, **Option A** is the correct answer.

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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If nP2=20⋅nP0\displaystyle ^nP_2 = 20 \cdot ^nP_0nP2​=20⋅nP0​, then the value of n\displaystyle nn is given by: