15C3r = 15Cr+3 , then ' r ' is equal to

Option b: 3. Solution: Given equation: 15C3r = 15Cr+3 We know the standard combinatorial property: If nCx = nCy , then either: 1. x = y , or 2. x + y = n Let us check both cases: Case 1: 3r = r + 3 2r = 3 r = 3/2 = 1.5 Since the subscript r in nCr must be a non-negative integer, r = 1.5 is not a valid solution. Case 2: 3r + (r + 3) = 15 4r + 3 = 15 4r = 12 r = 3 Since r = 3 is a positive integer, it is a valid solution. Let us verify by substituting r = 3 back into the original expression: LHS = 15C3(3) = 15C9 RHS = 15C3+3 = 15C6 Since 15C9 = 15!/9! × 6! and 15C6 = 15!/6! × 9! , LHS is indeed equal to RHS. Hence, Option B is the correct answer.

$^{15}C_{3r} = ^{15}C_{r+3}$, then '$r$' is equal to

The correct answer is Option b: 3. Given equation: $$^{15}C_{3r} = ^{15}C_{r+3}$$ We know the standard combinatorial property: If $^{n}C_{x} = ^{n}C_{y}$, then either: 1. $x = y$, or 2. $x + y = n$ Let us check both cases: **Case 1:** $3r = r + 3$ $$\implies 2r = 3 \implies r = \frac{3}{2} = 1.5$$ Since the subscript $r$ in $^{n}C_{r}$ must be a non-negative integer, $r = 1.5$ is not a valid solution. **Case 2:** $3r + (r + 3) = 15$ $$\implies 4r + 3 = 15$$ $$\implies 4r = 12$$ $$\implies r = 3$$ Since $r = 3$ is a positive integer, it is a valid solution. Let us verify by substituting $r = 3$ back into the original expression: $$\text{LHS} = ^{15}C_{3(3)} = ^{15}C_9$$ $$\text{RHS} = ^{15}C_{3+3} = ^{15}C_6$$ Since $^{15}C_9 = \frac{15!}{9! \times 6!}$ and $^{15}C_6 = \frac{15!}{6! \times 9!}$, LHS is indeed equal to RHS. Hence, **Option B** is the correct answer.

Permutations and CombinationsMCQMTP March 22Question 1714 of 251

All Questions

$\displaystyle ^{15}C_{3r} = ^{15}C_{r+3}$ , then ' $\displaystyle r$ ' is equal to

Options

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Correct Answer

✅ Option b — 3

All Options:

Detailed Solution & Explanation

Given equation:

^{15}C_{3r} = ^{15}C_{r+3}

We know the standard combinatorial property:
If

\displaystyle ^{n}C_{x} = ^{n}C_{y}

, then either:
1.

\displaystyle x = y

, or
2.

\displaystyle x + y = n

Let us check both cases:

**Case 1:**

\displaystyle 3r = r + 3

\implies 2r = 3 \implies r = \frac{3}{2} = 1.5

Since the subscript

\displaystyle r

\displaystyle ^{n}C_{r}

must be a non-negative integer,

\displaystyle r = 1.5

is not a valid solution.

**Case 2:**

\displaystyle 3r + (r + 3) = 15

\implies 4r + 3 = 15

\implies 4r = 12

\implies r = 3

Since

\displaystyle r = 3

is a positive integer, it is a valid solution.

Let us verify by substituting

\displaystyle r = 3

back into the original expression:

\text{LHS} = ^{15}C_{3(3)} = ^{15}C_9

\text{RHS} = ^{15}C_{3+3} = ^{15}C_6

Since

\displaystyle ^{15}C_9 = \frac{15!}{9! \times 6!}

and

\displaystyle ^{15}C_6 = \frac{15!}{6! \times 9!}

, LHS is indeed equal to RHS.

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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15C3r=15Cr+3\displaystyle ^{15}C_{3r} = ^{15}C_{r+3}15C3r​=15Cr+3​, then 'r\displaystyle rr' is equal to