In how many ways 3 prizes out of 5 can be distributed amongst 3 brothers equally

Option c: 60. Solution: We are required to distribute 3 prizes selected from a total of 5 distinct prizes among 3 brothers equally. "Equally" implies that each of the 3 brothers must receive exactly 1 prize. This process can be broken down into two sequential steps: - Step 1: Select 3 prizes out of 5 distinct prizes The number of ways to choose 3 prizes is given by 5C3 : 5C3 = 5!/3! × 2! = 5 × 4/2 × 1 = 10 - Step 2: Distribute the 3 selected prizes among the 3 brothers Since both the prizes and the brothers are distinct, the number of ways to assign 3 distinct prizes to 3 distinct brothers (one prize each) is: 3! = 3 × 2 × 1 = 6 ways - Step 3: Calculate the total ways By the multiplication principle, the total number of ways is: Total Ways = 5C3 × 3! = 10 × 6 = 60 Hence, Option C is the correct answer.

Permutations and CombinationsMCQMTP June 24 Series IIIQuestion 1747 of 251

All Questions

In how many ways $\displaystyle 3$ prizes out of $\displaystyle 5$ can be distributed amongst $\displaystyle 3$ brothers equally

Options

\displaystyle 10

\displaystyle 45

\displaystyle 60

\displaystyle 120

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Correct Answer

✅ Option c — $\displaystyle 60$

All Options:

A $\displaystyle 10$
B $\displaystyle 45$
C $\displaystyle 60$
D $\displaystyle 120$

Detailed Solution & Explanation

We are required to distribute

\displaystyle 3

prizes selected from a total of

\displaystyle 5

distinct prizes among

\displaystyle 3

brothers equally.

"Equally" implies that each of the

\displaystyle 3

brothers must receive exactly

\displaystyle 1

prize. This process can be broken down into two sequential steps:

- **Step 1: Select 3 prizes out of 5 distinct prizes**
The number of ways to choose

\displaystyle 3

prizes is given by

\displaystyle ^{5}C_{3}

^{5}C_{3} = \frac{5!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10

- **Step 2: Distribute the 3 selected prizes among the 3 brothers**
Since both the prizes and the brothers are distinct, the number of ways to assign

\displaystyle 3

distinct prizes to

\displaystyle 3

distinct brothers (one prize each) is:

3! = 3 \times 2 \times 1 = 6 \text{ ways}

- **Step 3: Calculate the total ways**
By the multiplication principle, the total number of ways is:

\text{Total Ways} = ^{5}C_{3} \times 3! = 10 \times 6 = 60

Hence, **Option C** 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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In how many ways 3\displaystyle 33 prizes out of 5\displaystyle 55 can be distributed amongst 3\displaystyle 33 brothers equally