Permutations and CombinationsMCQMTP Dec 23 - Series IQuestion 1672 of 251
All Questions

The value of N\displaystyle N in 17!+18!=N9!\displaystyle \frac{1}{7!} + \frac{1}{8!} = \frac{N}{9!} is

Options

A81
B78
C89
D64
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Correct Answer

✅ Option a — 81

All Options:

  • A81
  • B78
  • C89
  • D64

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Detailed Solution & Explanation

We are given the algebraic equation:
17!+18!=N9!\frac{1}{7!} + \frac{1}{8!} = \frac{N}{9!}

To solve for N\displaystyle N, let us express all factorials in terms of the smallest factorial present, which is 7!\displaystyle 7!.
We know that:
8!=8×7!8! = 8 \times 7!
9!=9×8×7!=72×7!9! = 9 \times 8 \times 7! = 72 \times 7!

Substituting these values into the given equation:
17!+18×7!=N72×7!\frac{1}{7!} + \frac{1}{8 \times 7!} = \frac{N}{72 \times 7!}

Now, we factor out 17!\displaystyle \frac{1}{7!} from both sides of the equation:
17!(1+18)=17!(N72)\frac{1}{7!} \left(1 + \frac{1}{8}\right) = \frac{1}{7!} \left(\frac{N}{72}\right)

Since 17!≠0\displaystyle \frac{1}{7!} \neq 0, we can divide both sides by 17!\displaystyle \frac{1}{7!}:
1+18=N721 + \frac{1}{8} = \frac{N}{72}
98=N72\frac{9}{8} = \frac{N}{72}

Solving for N\displaystyle N:
N=72×98=9×9=81N = 72 \times \frac{9}{8} = 9 \times 9 = 81

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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