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Q1(i):
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed?

Solution :

Step 1: Analyzing the Positional Constraints

We are required to form a 3-digit number using the set of digits $S = \{1, 2, 3, 4, 5\}$. A 3-digit number consists of three distinct positions: the Hundreds place, the Tens place, and the Units place.

Position Hundreds Place Tens Place Units Place
Number of Choices 5 5 5

Step 2: Applying the Fundamental Principle of Counting (Multiplication Principle)

Since the repetition of digits is allowed, the selection of a digit for one position does not restrict the choices for the subsequent positions. According to the Fundamental Principle of Counting, if an operation can be performed in $n$ ways, and a subsequent independent operation can be performed in $m$ ways, then the total number of ways to perform both is $n \times m$.

For a 3-digit number:

  • Number of ways to fill the Hundreds place ($n_1$): $5$
  • Number of ways to fill the Tens place ($n_2$): $5$
  • Number of ways to fill the Units place ($n_3$): $5$

Step 3: Mathematical Calculation

The total number of 3-digit numbers is given by the product of the number of choices for each position:

$Total = n_1 \times n_2 \times n_3$

$Total = 5 \times 5 \times 5$

$Total = 5^3$

$Total = 125$

5 5 5 Hundreds Tens Units

Final Solution: The total number of 3-digit numbers that can be formed is 125.


More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.1


CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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