Permutations and CombinationsMCQMTP May 19Question 1662 of 251
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The number of ways in which 8 examination papers be arranged so that the best and worst papers never come together

Options

A8!−2×7!\displaystyle 8! - 2 \times 7!
B8!−7!\displaystyle 8! - 7!
C8!\displaystyle 8!
D7!\displaystyle 7!
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Correct Answer

✅ Option a — 8!−2×7!\displaystyle 8! - 2 \times 7!

All Options:

  • A8!−2×7!\displaystyle 8! - 2 \times 7!
  • B8!−7!\displaystyle 8! - 7!
  • C8!\displaystyle 8!
  • D7!\displaystyle 7!

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Detailed Solution & Explanation

To find the number of ways to arrange the 8 examination papers such that the best and worst papers never come together, we can use the method of complementation.

First, let us calculate the total number of unrestricted arrangements of the 8 papers. Since all 8 papers are distinct, the number of ways to arrange them in a row is given by the permutation of 8 objects:
Total arrangements=8!\text{Total arrangements} = 8!

Second, let us calculate the number of arrangements in which the best paper (B\displaystyle B) and the worst paper (W\displaystyle W) are always together.
We can treat the best and worst papers as a single compound object or block: [B,W]\displaystyle [B, W].
Now, we have this 1 block and the remaining 8−2=6\displaystyle 8 - 2 = 6 individual papers. This gives us a total of:
1+6=7 objects to arrange.1 + 6 = 7 \text{ objects to arrange.}
The number of ways to arrange these 7 distinct objects in a row is 7!\displaystyle 7!.
Within the block [B,W]\displaystyle [B, W], the best and worst papers can be arranged among themselves in 2!=2\displaystyle 2! = 2 ways (either as BW\displaystyle BW or WB\displaystyle WB).
Thus, the number of arrangements where the best and worst papers are always together is:
Arrangements together=2×7!\text{Arrangements together} = 2 \times 7!

Third, the number of ways in which the best and worst papers never come together is the difference between the total unrestricted arrangements and the arrangements where they are together:
Arrangements never together=Total arrangements−Arrangements together\text{Arrangements never together} = \text{Total arrangements} - \text{Arrangements together}
Arrangements never together=8!−2×7!\text{Arrangements never together} = 8! - 2 \times 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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