Permutations and CombinationsMCQMTP Sep 24 Series IIQuestion 1756 of 251
All Questions

If 2nC2:nC2=11:1\displaystyle ^{2n}C_2 : ^{n}C_2 = 11:1 the value of n\displaystyle n is

Options

A6\displaystyle 6
B7\displaystyle 7
C5\displaystyle 5
DNone of these
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Correct Answer

✅ Option a — 6\displaystyle 6

All Options:

  • A6\displaystyle 6
  • B7\displaystyle 7
  • C5\displaystyle 5
  • DNone of these

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Detailed Solution & Explanation

To find the value of n\displaystyle n, let us analyze the given relationship:
Given ratio: 2nC2:nC2=11:1\text{Given ratio: } ^{2n}C_2 : ^nC_2 = 11 : 1
We know the combination formula is:
mCr=m!r!(m−r)!^mC_r = \frac{m!}{r!(m-r)!}
Using this formula, we can express the two terms as:
2nC2=(2n)!2!(2n−2)!=2n(2n−1)2=n(2n−1)^{2n}C_2 = \frac{(2n)!}{2!(2n-2)!} = \frac{2n(2n-1)}{2} = n(2n-1)
nC2=n!2!(n−2)!=n(n−1)2^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}
Now, substituting these into the given ratio:
2nC2nC2=n(2n−1)n(n−1)2=11\frac{^{2n}C_2}{^nC_2} = \frac{n(2n-1)}{\frac{n(n-1)}{2}} = 11
2n(2n−1)n(n−1)=11\frac{2n(2n-1)}{n(n-1)} = 11
Since n≥2\displaystyle n \ge 2, we can divide both numerator and denominator by n\displaystyle n:
2(2n−1)n−1=11\frac{2(2n-1)}{n-1} = 11
2(2n−1)=11(n−1)2(2n-1) = 11(n-1)
4n−2=11n−114n - 2 = 11n - 11
11n−4n=11−211n - 4n = 11 - 2
7n=9  ⟹  n=977n = 9 \implies n = \frac{9}{7}
Since n\displaystyle n must be a positive integer (n∈Z+\displaystyle n \in \mathbb{Z}^+), a value of n=97\displaystyle n = \frac{9}{7} is not mathematically possible for combinations. This suggests there is a typographical error in the question as printed in the model test paper, where 2nC3:nC3=11:1\displaystyle ^{2n}C_3 : ^nC_3 = 11:1 was likely intended instead of 2nC2:nC2=11:1\displaystyle ^{2n}C_2 : ^nC_2 = 11:1.

Let us solve the intended standard mathematical problem:
Intended ratio: 2nC3:nC3=11:1\text{Intended ratio: } ^{2n}C_3 : ^nC_3 = 11 : 1
Using the combination formula:
2nC3=(2n)!3!(2n−3)!=2n(2n−1)(2n−2)6=2n(2n−1)⋅2(n−1)6=2n(2n−1)(n−1)3^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{2n(2n-1)(2n-2)}{6} = \frac{2n(2n-1) \cdot 2(n-1)}{6} = \frac{2n(2n-1)(n-1)}{3}
nC3=n!3!(n−3)!=n(n−1)(n−2)6^nC_3 = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{6}
Now, substituting these expressions into the ratio:
2nC3nC3=2n(2n−1)(n−1)3n(n−1)(n−2)6=11\frac{^{2n}C_3}{^nC_3} = \frac{\frac{2n(2n-1)(n-1)}{3}}{\frac{n(n-1)(n-2)}{6}} = 11
2n(2n−1)(n−1)3×6n(n−1)(n−2)=11\frac{2n(2n-1)(n-1)}{3} \times \frac{6}{n(n-1)(n-2)} = 11
For n≥3\displaystyle n \ge 3, the terms n\displaystyle n and (n−1)\displaystyle (n-1) in the numerator and denominator cancel out:
2(2n−1)⋅2n−2=11\frac{2(2n-1) \cdot 2}{n-2} = 11
4(2n−1)n−2=11\frac{4(2n-1)}{n-2} = 11
4(2n−1)=11(n−2)4(2n-1) = 11(n-2)
8n−4=11n−228n - 4 = 11n - 22
11n−8n=22−411n - 8n = 22 - 4
3n=18  ⟹  n=63n = 18 \implies n = 6
This perfectly matches **Option A** (6\displaystyle 6).

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.

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Permutations vs Combinations: The Quantitative Aptitude Guide

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