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Q1:
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Solution :

Step 1: Establishing the Mathematical Framework

To determine the number of 3-digit numbers that can be formed using the set of digits $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ without repetition, we employ the Fundamental Principle of Counting, specifically the Permutation formula. Since the order of digits is essential for forming unique numbers, we are selecting and arranging $r$ objects from a set of $n$ distinct objects.

The formula for permutations of $n$ distinct objects taken $r$ at a time is given by:

$P(n, r) = \frac{n!}{(n-r)!}$

[Per the Fundamental Counting Principle and the definition of Permutations]

Step 2: Identifying Given Variables

  • Total number of distinct digits available ($n$): $9$
  • Number of positions to fill (3-digit number, so $r$): $3$
  • Constraint: No digit is repeated (sampling without replacement).

Step 3: Execution of the Permutation Calculation

Substituting the values into the formula:

$P(9, 3) = \frac{9!}{(9-3)!}$

$P(9, 3) = \frac{9!}{6!}$

Expanding the factorial terms to simplify:

$P(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{6!}$

$P(9, 3) = 9 \times 8 \times 7$

$P(9, 3) = 72 \times 7 = 504$

Step 4: Logical Verification via Multiplication Principle

Alternatively, using the "slot method" for independent events:

Position Available Choices
Hundreds Place 9 options
Tens Place 8 options (9 total minus the digit used in hundreds place)
Units Place 7 options (9 total minus the two digits used previously)

Total arrangements = $9 \times 8 \times 7 = 504$. [By the Multiplication Principle of Counting]

9 8 7 Ă— Ă— Hundreds Tens Units

Final Solution: The total number of 3-digit numbers that can be formed is 504.


More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.3


CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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