Permutations and CombinationsMCQPYQ Nov. 18Question 1598 of 251
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The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} 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

To find the value of N\displaystyle N, let us first examine the equation given in the question:
N!+frac17!=frac18!N! + \\frac{1}{7!} = \\frac{1}{8!}
However, solving this directly for N!\displaystyle N! yields:
N!=frac18!−frac17!=frac18!−frac88!=−frac78!N! = \\frac{1}{8!} - \\frac{1}{7!} = \\frac{1}{8!} - \\frac{8}{8!} = -\\frac{7}{8!}
Since a factorial cannot be negative, this equation is mathematically impossible and indicates a typographical error in the question text.

The standard CA Foundation question of this type is:
frac17!+frac18!=fracN9!\\frac{1}{7!} + \\frac{1}{8!} = \\frac{N}{9!}
Let us solve this corrected version step-by-step:
frac17!+frac18times7!=fracN9times8times7!\\frac{1}{7!} + \\frac{1}{8 \\times 7!} = \\frac{N}{9 \\times 8 \\times 7!}
Multiplying the entire equation by 7!\displaystyle 7! to clear the denominators:
1+frac18=fracN721 + \\frac{1}{8} = \\frac{N}{72}
frac98=fracN72\\frac{9}{8} = \\frac{N}{72}
N=frac98times72N = \\frac{9}{8} \\times 72
N=9times9=81N = 9 \\times 9 = 81
This perfectly matches the value of Option A (81\displaystyle 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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If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

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