Permutations and CombinationsPYQ Jan 26Question 4264 of 183
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How many 3-digit number can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digit is repeated?

Options

A18
B20
C22
D24
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Correct Answer

✅ Option b — 20

All Options:

  • A18
  • B20
  • C22
  • D24

Detailed Solution & Explanation

We are given the digits: 2,3,5,6,7,9\displaystyle 2, 3, 5, 6, 7, 9 (total of 6\displaystyle 6 digits) with no repetition allowed. We want to form a 3\displaystyle 3-digit number that is divisible by 5\displaystyle 5.
For a number to be divisible by 5\displaystyle 5, its units digit must be 0\displaystyle 0 or 5\displaystyle 5. Since the digit 0\displaystyle 0 is not in our list, the units place must be filled by the digit 5\displaystyle 5: - Number of ways to fill the units place = 1\displaystyle 1 (using digit 5\displaystyle 5)
Since repetition is not allowed, the remaining two places (hundreds and tens) can be filled using the remaining 5\displaystyle 5 digits (2,3,6,7,9\displaystyle 2, 3, 6, 7, 9) in 5P2\displaystyle ^5P_2 ways: 5P2=5×4=20 ways^5P_2 = 5 \times 4 = 20\text{ ways}
Thus, the total number of such 3\displaystyle 3-digit numbers is: Total=20×1=20\text{Total} = 20 \times 1 = 20 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.

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