Permutations and CombinationsMCQMTP June 24 Series IIQuestion 1745 of 251
All Questions

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all the five balls. In how many different ways can we place the balls so that no box remains empty?

Options

A100\displaystyle 100
B120\displaystyle 120
C150\displaystyle 150
DNone of these
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Correct Answer

✅ Option c — 150\displaystyle 150

All Options:

  • A100\displaystyle 100
  • B120\displaystyle 120
  • C150\displaystyle 150
  • DNone of these

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Detailed Solution & Explanation

We are required to place 5\displaystyle 5 distinct balls of different colors into 3\displaystyle 3 distinct boxes of different sizes such that **no box remains empty**.

This is equivalent to finding the number of onto functions (surjective distributions) from a set of 5\displaystyle 5 elements to a set of 3\displaystyle 3 elements. We can solve this using the **Principle of Inclusion-Exclusion (PIE)**:

Let N\displaystyle N be the total number of ways to distribute 5\displaystyle 5 distinct balls into 3\displaystyle 3 distinct boxes without any restrictions:
N=35=243 waysN = 3^5 = 243 \text{ ways}
Let us define the conditions:
- A\displaystyle A: Box 1 is empty.
- B\displaystyle B: Box 2 is empty.
- C\displaystyle C: Box 3 is empty.

The number of ways where at least one box is empty is:
Empty Ways=∑N(A)−∑N(A∩B)+N(A∩B∩C)\text{Empty Ways} = \sum N(A) - \sum N(A \cap B) + N(A \cap B \cap C)
- **Step 1: Calculate ∑N(A)\displaystyle \sum N(A) (number of ways where 1 specific box is empty):**
If one specific box is empty, the balls must be placed in the other 2\displaystyle 2 boxes.
N(A)=25=32 waysN(A) = 2^5 = 32 \text{ ways}
Since there are 3\displaystyle 3 boxes, we choose 1\displaystyle 1 box to be empty in 3C1=3\displaystyle ^{3}C_{1} = 3 ways.
∑N(A)=3×32=96 ways\sum N(A) = 3 \times 32 = 96 \text{ ways}
- **Step 2: Calculate ∑N(A∩B)\displaystyle \sum N(A \cap B) (number of ways where 2 specific boxes are empty):**
If two specific boxes are empty, all 5\displaystyle 5 balls must be placed in the single remaining box.
N(A∩B)=15=1 wayN(A \cap B) = 1^5 = 1 \text{ way}
We choose 2\displaystyle 2 boxes to be empty in 3C2=3\displaystyle ^{3}C_{2} = 3 ways.
∑N(A∩B)=3×1=3 ways\sum N(A \cap B) = 3 \times 1 = 3 \text{ ways}
- **Step 3: Calculate N(A∩B∩C)\displaystyle N(A \cap B \cap C) (all 3 boxes are empty):**
It is impossible to distribute 5\displaystyle 5 balls such that all 3\displaystyle 3 boxes are empty.
N(A∩B∩C)=0 waysN(A \cap B \cap C) = 0 \text{ ways}
- **Step 4: Applying Inclusion-Exclusion:**
Ways with no empty box=Total Ways−(∑N(A)−∑N(A∩B)+N(A∩B∩C))\text{Ways with no empty box} = \text{Total Ways} - \left(\sum N(A) - \sum N(A \cap B) + N(A \cap B \cap C)\right)
Ways with no empty box=243−(96−3+0)=243−93=150 ways\text{Ways with no empty box} = 243 - (96 - 3 + 0) = 243 - 93 = 150 \text{ ways}

**Discrepancy Note:**
The mathematical solution is exactly 150\displaystyle 150 ways, which corresponds to **Option C**. The textbook answer key has a typographical error, incorrectly indicating **Option D** (None of these) as the correct answer.

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.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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