Permutations and CombinationsMCQMTP March 22Question 1733 of 251
All Questions

X and Y stand in a line with 6 other people. What is the probability that there are 3 persons between them?

Options

A15\displaystyle \frac{1}{5}
B16\displaystyle \frac{1}{6}
C17\displaystyle \frac{1}{7}
D13\displaystyle \frac{1}{3}
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Correct Answer

✅ Option c — 17\displaystyle \frac{1}{7}

All Options:

  • A15\displaystyle \frac{1}{5}
  • B16\displaystyle \frac{1}{6}
  • C17\displaystyle \frac{1}{7}
  • D13\displaystyle \frac{1}{3}

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Detailed Solution & Explanation

The total number of people standing in the line is 2 (X and Y)+6 others=8\displaystyle 2 \text{ (X and Y)} + 6 \text{ others} = 8 people.

The total number of ways to arrange 8\displaystyle 8 distinct people in a line is:
Total Arrangements=8!\text{Total Arrangements} = 8!
Now we find the number of favorable arrangements where there are **exactly 3 people between X and Y**.
Let the 8\displaystyle 8 positions in the line be labeled 1,2,3,4,5,6,7,8\displaystyle 1, 2, 3, 4, 5, 6, 7, 8. For there to be exactly 3\displaystyle 3 people between X and Y, the difference between their positions must be exactly 4\displaystyle 4 (i.e., ∣i−j∣=4\displaystyle |i - j| = 4). The possible pairs of positions for X and Y are:
- (1,5)\displaystyle (1, 5)
- (2,6)\displaystyle (2, 6)
- (3,7)\displaystyle (3, 7)
- (4,8)\displaystyle (4, 8)

Since X and Y can be interchanged in each pair (e.g., X at 1\displaystyle 1 and Y at 5\displaystyle 5, or Y at 1\displaystyle 1 and X at 5\displaystyle 5), there are:
4×2=8 ways to position X and Y4 \times 2 = 8 \text{ ways to position X and Y}
For each of these 8\displaystyle 8 positional choices, the remaining 6\displaystyle 6 people can be arranged in the remaining 6\displaystyle 6 empty spots in:
6! ways6! \text{ ways}
Thus, the total number of favorable arrangements is:
Favorable Arrangements=8×6!\text{Favorable Arrangements} = 8 \times 6!
The probability P\displaystyle P of this event is the ratio of favorable arrangements to total arrangements:
P=Favorable ArrangementsTotal Arrangements=8×6!8!P = \frac{\text{Favorable Arrangements}}{\text{Total Arrangements}} = \frac{8 \times 6!}{8!}
  ⟹  P=8×6!8×7×6!=17\implies P = \frac{8 \times 6!}{8 \times 7 \times 6!} = \frac{1}{7}

Hence, **Option C** 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.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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

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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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