Permutations and CombinationsMCQMTP June 22Question 1715 of 251
All Questions

The number in ways in which 4 persons can occupy 9 vacant seats is

Options

A6048
B3024
C1512
D4536
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Correct Answer

✅ Option b — 3024

All Options:

  • A6048
  • B3024
  • C1512
  • D4536

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Detailed Solution & Explanation

We are required to find the number of ways in which 4\displaystyle 4 distinct persons can occupy 9\displaystyle 9 vacant seats.

Since the persons are distinct and the seats are distinct, the order of seating matters. This is a problem of permutations, specifically choosing and arranging 4\displaystyle 4 items out of 9\displaystyle 9.

The number of ways is given by the formula for permutations:
nPr=n!(n−r)!^nP_r = \frac{n!}{(n-r)!}
Here, n=9\displaystyle n = 9 (total vacant seats) and r=4\displaystyle r = 4 (number of persons to be seated).

Substituting these values:
9P4=9!(9−4)!=9!5!^9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!}
  ⟹  9P4=9×8×7×6×5!5!\implies ^9P_4 = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!}
  ⟹  9P4=9×8×7×6\implies ^9P_4 = 9 \times 8 \times 7 \times 6
  ⟹  9P4=72×42=3024\implies ^9P_4 = 72 \times 42 = 3024

Alternatively, this can be understood step-by-step:
- The first person has 9\displaystyle 9 choices of seats.
- The second person has 8\displaystyle 8 choices remaining.
- The third person has 7\displaystyle 7 choices remaining.
- The fourth person has 6\displaystyle 6 choices remaining.

By the multiplication principle, the total number of ways is:
Total Ways=9×8×7×6=3024\text{Total Ways} = 9 \times 8 \times 7 \times 6 = 3024

Hence, **Option B** 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

More Questions from Permutations and Combinations

The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} is

A person can go from place 'A' to 'B' by 11 different modes of transport but is allowed to return to 'A' by any mode other than the one earlier. The number of different ways in which the entire journey can be completed is:

If a man travels from place A to B in 10 ways then by how many ways can he come back by another train?

If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

Which of the following is a correct statement.

nP3=2⋅nP2\displaystyle ^nP_3 = 2 \cdot ^nP_2. Find n\displaystyle n.

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