Permutations and CombinationsMCQPYQ June 19Question 1603 of 251
All Questions

Which of the following is a correct statement.

Options

AnPn=nPn−1\displaystyle ^nP_n = ^nP_{n-1}
BnPn=2nPn−2\displaystyle ^nP_n = 2n P_{n-2}
CnPn=n!nPn−1\displaystyle ^nP_n = \frac{n!}{n P_{n-1}}
DnPn=nPn−1n\displaystyle ^nP_n = ^nP_{n-1} n
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Correct Answer

✅ Option a — nPn=nPn−1\displaystyle ^nP_n = ^nP_{n-1}

All Options:

  • AnPn=nPn−1\displaystyle ^nP_n = ^nP_{n-1}
  • BnPn=2nPn−2\displaystyle ^nP_n = 2n P_{n-2}
  • CnPn=n!nPn−1\displaystyle ^nP_n = \frac{n!}{n P_{n-1}}
  • DnPn=nPn−1n\displaystyle ^nP_n = ^nP_{n-1} n

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Detailed Solution & Explanation

Let us analyze the formulas for each option using the definition of permutations:
The standard permutation formula is:
nPr=fracn!(n−r)!^nP_r = \\frac{n!}{(n-r)!}
1. **Evaluating nPn\displaystyle ^nP_n:**
nPn=fracn!(n−n)!=fracn!0!=fracn!1=n!^nP_n = \\frac{n!}{(n-n)!} = \\frac{n!}{0!} = \\frac{n!}{1} = n!
2. **Evaluating nPn−1\displaystyle ^nP_{n-1}:**
nPn−1=fracn!(n−(n−1))!=fracn!1!=n!^nP_{n-1} = \\frac{n!}{(n-(n-1))!} = \\frac{n!}{1!} = n!
Comparing the two results, we get:
nPn=nPn−1^nP_n = ^nP_{n-1}
This shows that statement (a) is mathematically correct.
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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