If nC2 = nC3 , the value of 'n' is

Option d: None of these. Solution: Given equation: nC2 = nC3 We use the standard property of combinations: If nCx = nCy , then either: 1. x = y , or 2. x + y = n Since 2 ≠ 3 , the first case is not possible. Therefore, we must have: 2 + 3 = n n = 5 Let us verify by substituting n = 5 back into the original equation: LHS = 5C2 = 5 × 4/2 × 1 = 10 RHS = 5C3 = 5 × 4 × 3/3 × 2 × 1 = 10 Since LHS = RHS = 10 , the solution n = 5 is correct. Since the value n = 5 is not listed in the options : 10, B: 14, C: 12 , the mathematically correct choice is Option D (None of these). Discrepancy Note: The textbook answer key has a typographical error, incorrectly indicating Option C ( 12 ) as the correct answer. This is mathematically incorrect, as 12C2 = 66 and 12C3 = 220 , which are not equal. Hence, Option D is the correct answer.

If $^{n}C_2 = ^{n}C_3$, the value of 'n' is

The correct answer is Option d: None of these. Given equation: $$^{n}C_2 = ^{n}C_3$$ We use the standard property of combinations: If $^{n}C_x = ^{n}C_y$, then either: 1. $x = y$, or 2. $x + y = n$ Since $2 \neq 3$, the first case is not possible. Therefore, we must have: $$2 + 3 = n$$ $$\implies n = 5$$ Let us verify by substituting $n = 5$ back into the original equation: $$\text{LHS} = ^{5}C_2 = \frac{5 \times 4}{2 \times 1} = 10$$ $$\text{RHS} = ^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$ Since LHS = RHS = $10$, the solution $n = 5$ is correct. Since the value $n = 5$ is not listed in the options $\{\text{A: } 10, \text{B: } 14, \text{C: } 12\}$, the mathematically correct choice is **Option D (None of these)**. **Discrepancy Note:** The textbook answer key has a typographical error, incorrectly indicating **Option C** ($12$) as the correct answer. This is mathematically incorrect, as $^{12}C_2 = 66$ and $^{12}C_3 = 220$, which are not equal. Hence, **Option D** is the correct answer.

Permutations and CombinationsMCQMTP Sep 24 Series IIQuestion 1755 of 251

All Questions

If $\displaystyle ^{n}C_2 = ^{n}C_3$ , the value of 'n' is

Options

\displaystyle 10

\displaystyle 14

\displaystyle 12

DNone of these

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option d — None of these

All Options:

A $\displaystyle 10$
B $\displaystyle 14$
C $\displaystyle 12$
DNone of these

Detailed Solution & Explanation

Given equation:

^{n}C_2 = ^{n}C_3

We use the standard property of combinations:
If

\displaystyle ^{n}C_x = ^{n}C_y

, then either:
1.

\displaystyle x = y

, or
2.

\displaystyle x + y = n

Since

\displaystyle 2 \neq 3

, the first case is not possible. Therefore, we must have:

2 + 3 = n

\implies n = 5

Let us verify by substituting

\displaystyle n = 5

back into the original equation:

\text{LHS} = ^{5}C_2 = \frac{5 \times 4}{2 \times 1} = 10

\text{RHS} = ^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

Since LHS = RHS =

\displaystyle 10

, the solution

\displaystyle n = 5

is correct.

Since the value

\displaystyle n = 5

is not listed in the options

\displaystyle \{\text{A: } 10, \text{B: } 14, \text{C: } 12\}

, the mathematically correct choice is **Option D (None of these)**.

**Discrepancy Note:**
The textbook answer key has a typographical error, incorrectly indicating **Option C** (

\displaystyle 12

) as the correct answer. This is mathematically incorrect, as

\displaystyle ^{12}C_2 = 66

and

\displaystyle ^{12}C_3 = 220

, which are not equal.

Hence, **Option D** 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

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

If nC2=nC3\displaystyle ^{n}C_2 = ^{n}C_3nC2​=nC3​, the value of 'n' is