If $^nP_4 = 20 ^nP_2$ then the value of 'n' is _______.

The correct answer is Option a: 7. Given the equation: $$^nP_4 = 20 \cdot ^nP_2$$ Using the definition of permutations $^nP_k = \frac{n!}{(n-k)!}$: $$\frac{n!}{(n-4)!} = 20 \cdot \frac{n!}{(n-2)!}$$ Assuming $n \ge 4$ (so $n! \ne 0$): $$\frac{1}{(n-4)!} = \frac{20}{(n-2)(n-3)(n-4)!}$$ Multiplying both sides by $(n-4)!$: $$1 = \frac{20}{(n-2)(n-3)}$$ $$(n-2)(n-3) = 20$$ $$n^2 - 5n + 6 = 20$$ $$n^2 - 5n - 14 = 0$$ Factoring the quadratic equation: $$(n-7)(n+2) = 0$$ Since $n$ represents a positive number of items and must satisfy $n \ge 4$ for $^nP_4$ to be defined, $n = -2$ is rejected. Thus, $n = 7$. Hence, **Option A** is the correct answer.

Permutations and CombinationsMCQMTP Mar 22Question 1642 of 251

All Questions

If $\displaystyle ^nP_4 = 20 ^nP_2$ then the value of 'n' is _______.

Options

C-2 and 7 both

Dnone of these.

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Correct Answer

✅ Option a — 7

All Options:

A7
B7
C-2 and 7 both
Dnone of these.

Detailed Solution & Explanation

Given the equation:

^nP_4 = 20 \cdot ^nP_2

Using the definition of permutations

\displaystyle ^nP_k = \frac{n!}{(n-k)!}

\frac{n!}{(n-4)!} = 20 \cdot \frac{n!}{(n-2)!}

Assuming

\displaystyle n \ge 4

(so

\displaystyle n! \ne 0

\frac{1}{(n-4)!} = \frac{20}{(n-2)(n-3)(n-4)!}

Multiplying both sides by

\displaystyle (n-4)!

1 = \frac{20}{(n-2)(n-3)}

(n-2)(n-3) = 20

n^2 - 5n + 6 = 20

n^2 - 5n - 14 = 0

Factoring the quadratic equation:

(n-7)(n+2) = 0

Since

\displaystyle n

represents a positive number of items and must satisfy

\displaystyle n \ge 4

for

\displaystyle ^nP_4

to be defined,

\displaystyle n = -2

is rejected.
Thus,

\displaystyle n = 7

.
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 nP4=20nP2\displaystyle ^nP_4 = 20 ^nP_2nP4​=20nP2​ then the value of 'n' is _______.