Permutations and CombinationsMCQPYQ June 22Question 1602 of 251
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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'.

Options

A4
B5
C6
D7
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Correct Answer

✅ Option b — 5

All Options:

  • A4
  • B5
  • C6
  • D7

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Detailed Solution & Explanation

Let us first solve the equation exactly as printed in the question:
fracn!10(n−1)!=frac(n−1)!(n−3)!\\frac{n!}{10(n-1)!} = \\frac{(n-1)!}{(n-3)!}
Using the definition of factorials, we know that n!=ntimes(n−1)!\displaystyle n! = n \\times (n-1)! and (n−1)!=(n−1)(n−2)(n−3)!\displaystyle (n-1)! = (n-1)(n-2)(n-3)!. Substituting these into the equation:
fracntimes(n−1)!10(n−1)!=frac(n−1)(n−2)(n−3)!(n−3)!\\frac{n \\times (n-1)!}{10(n-1)!} = \\frac{(n-1)(n-2)(n-3)!}{(n-3)!}
Canceling the common factorial terms on both sides:
fracn10=(n−1)(n−2)\\frac{n}{10} = (n-1)(n-2)
n=10(n2−3n+2)n = 10(n^2 - 3n + 2)
n=10n2−30n+20n = 10n^2 - 30n + 20
10n2−31n+20=010n^2 - 31n + 20 = 0
Solving this quadratic equation yields non-integer values for n\displaystyle n (n=2.5\displaystyle n = 2.5 or n=0.8\displaystyle n = 0.8), which are invalid for factorials and do not match any options.

This indicates a typographical error in the question. The intended equation is:
fracn!10(n−4)!=frac(n−1)!(n−3)!\\frac{n!}{10(n-4)!} = \\frac{(n-1)!}{(n-3)!}
Let us solve this intended equation step-by-step:
fracn(n−1)(n−2)(n−3)(n−4)!10(n−4)!=frac(n−1)(n−2)(n−3)!(n−3)!\\frac{n(n-1)(n-2)(n-3)(n-4)!}{10(n-4)!} = \\frac{(n-1)(n-2)(n-3)!}{(n-3)!}
Canceling the factorial terms:
fracn(n−1)(n−2)(n−3)10=(n−1)(n−2)\\frac{n(n-1)(n-2)(n-3)}{10} = (n-1)(n-2)
Assuming ngeq4\displaystyle n \\geq 4, we can divide both sides by (n−1)(n−2)\displaystyle (n-1)(n-2):
fracn(n−3)10=1\\frac{n(n-3)}{10} = 1
n(n−3)=10n(n-3) = 10
n2−3n−10=0n^2 - 3n - 10 = 0
(n−5)(n+2)=0(n-5)(n+2) = 0
Since n\displaystyle n must be a positive integer, we discard n=−2\displaystyle n = -2, leaving:
n=5n = 5
This perfectly matches Option B.
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

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