Permutations and CombinationsMCQMTP June 22Question 1646 of 251
All Questions

Find the number of even numbers greater than 100 that can be formed with the digits 0,1,2,3.

Options

A10
B15
C20
Dnone of these
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Correct Answer

✅ Option c — 20

All Options:

  • A10
  • B15
  • C20
  • Dnone of these

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Detailed Solution & Explanation

We need to find the number of even numbers greater than 100 that can be formed using the digits {0,1,2,3}\displaystyle \{0, 1, 2, 3\} without repetition.
Since the numbers must be greater than 100 and we have 4 distinct digits, the numbers can be either 3-digit or 4-digit numbers.
For a number to be even, its units digit must be 0\displaystyle 0 or 2\displaystyle 2.

**1. 3-Digit Even Numbers (> 100)**
Let the number be d1d2d3\displaystyle d_1 d_2 d_3. Since it is >100\displaystyle > 100, d1∈{1,2,3}\displaystyle d_1 \in \{1, 2, 3\} (no leading zero).
- **Case 1.1: Units digit d3=0\displaystyle d_3 = 0**
The first digit d1\displaystyle d_1 can be chosen from {1,2,3}\displaystyle \{1, 2, 3\} (3 options).
The tens digit d2\displaystyle d_2 can be chosen from the remaining 2 digits (2 options).
Number of ways = 3×2=6\displaystyle 3 \times 2 = 6 ways (these are 120, 130, 210, 230, 310, 320).
- **Case 1.2: Units digit d3=2\displaystyle d_3 = 2**
The first digit d1\displaystyle d_1 can be chosen from {1,3}\displaystyle \{1, 3\} (2 options, as 0 cannot be the first digit).
The tens digit d2\displaystyle d_2 can be chosen from the remaining 2 digits (2 options).
Number of ways = 2×2=4\displaystyle 2 \times 2 = 4 ways (these are 102, 132, 302, 312).
Total 3-digit even numbers = 6+4=10\displaystyle 6 + 4 = 10.

**2. 4-Digit Even Numbers**
Let the number be d1d2d3d4\displaystyle d_1 d_2 d_3 d_4. Since it is a 4-digit number, d1∈{1,2,3}\displaystyle d_1 \in \{1, 2, 3\}.
- **Case 2.1: Units digit d4=0\displaystyle d_4 = 0**
d1\displaystyle d_1 can be chosen from {1,2,3}\displaystyle \{1, 2, 3\} (3 options).
d2\displaystyle d_2 can be chosen from the remaining 2 digits (2 options).
d3\displaystyle d_3 can be chosen from the remaining 1 digit (1 option).
Number of ways = 3×2×1=6\displaystyle 3 \times 2 \times 1 = 6 ways.
- **Case 2.2: Units digit d4=2\displaystyle d_4 = 2**
d1\displaystyle d_1 can be chosen from {1,3}\displaystyle \{1, 3\} (2 options).
d2\displaystyle d_2 can be chosen from the remaining 2 digits (2 options).
d3\displaystyle d_3 can be chosen from the remaining 1 digit (1 option).
Number of ways = 2×2×1=4\displaystyle 2 \times 2 \times 1 = 4 ways.
Total 4-digit even numbers = 6+4=10\displaystyle 6 + 4 = 10.

Summing both cases, the total number of even numbers greater than 100 is:
Total=10+10=20\text{Total} = 10 + 10 = 20
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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