Question:
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When forming numbers with constraints like "odd" or "even", always start by filling the place that has the constraint first (in this case, the units place). Then fill the remaining places. The problem is simplified by noticing that the number of available odd digits is the same as the number of available even digits.
Updated On: Oct 3, 2025
- The quantity on the left is greater
- The quantity on the right is greater
- Both are equal
- The relationship cannot be determined without further information
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
This question involves forming three-digit numbers with specific constraints (odd or even) from a given set of digits, with repetition allowed. The key is to analyze the choices available for each digit's place (hundreds, tens, units), especially the units place which determines if a number is odd or even.
Step 2: Key Formula or Approach:
We will use the fundamental principle of counting. The total number of possibilities is the product of the number of choices for each of the three places.
Step 3: Detailed Explanation:
The set of available digits is {1, 4, 5, 6, 7, 8}. Total number of available digits is 6.
Repetition is allowed in both cases.
For Column A: Three-digit ODD numbers
A number is odd if its units digit is an odd number.
From the given set, the odd digits are {1, 5, 7}. There are 3 choices for the units place.
- Units place: 3 choices (1, 5, or 7).
- Tens place: 6 choices (any of the 6 digits, as repetition is allowed).
- Hundreds place: 6 choices (any of the 6 digits).
Total number of odd numbers = (Choices for hundreds) × (Choices for tens) × (Choices for units)
Total odd numbers = 6 × 6 × 3 = 108
For Column B: Three-digit EVEN numbers
A number is even if its units digit is an even number.
From the given set, the even digits are {4, 6, 8}. There are 3 choices for the units place.
- Units place: 3 choices (4, 6, or 8).
- Tens place: 6 choices (any of the 6 digits).
- Hundreds place: 6 choices (any of the 6 digits).
Total number of even numbers = (Choices for hundreds) × (Choices for tens) × (Choices for units)
Total even numbers = 6 × 6 × 3 = 108
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 108
Quantity B = 108
The two quantities are equal.
This question involves forming three-digit numbers with specific constraints (odd or even) from a given set of digits, with repetition allowed. The key is to analyze the choices available for each digit's place (hundreds, tens, units), especially the units place which determines if a number is odd or even.
Step 2: Key Formula or Approach:
We will use the fundamental principle of counting. The total number of possibilities is the product of the number of choices for each of the three places.
Step 3: Detailed Explanation:
The set of available digits is {1, 4, 5, 6, 7, 8}. Total number of available digits is 6.
Repetition is allowed in both cases.
For Column A: Three-digit ODD numbers
A number is odd if its units digit is an odd number.
From the given set, the odd digits are {1, 5, 7}. There are 3 choices for the units place.
- Units place: 3 choices (1, 5, or 7).
- Tens place: 6 choices (any of the 6 digits, as repetition is allowed).
- Hundreds place: 6 choices (any of the 6 digits).
Total number of odd numbers = (Choices for hundreds) × (Choices for tens) × (Choices for units)
Total odd numbers = 6 × 6 × 3 = 108
For Column B: Three-digit EVEN numbers
A number is even if its units digit is an even number.
From the given set, the even digits are {4, 6, 8}. There are 3 choices for the units place.
- Units place: 3 choices (4, 6, or 8).
- Tens place: 6 choices (any of the 6 digits).
- Hundreds place: 6 choices (any of the 6 digits).
Total number of even numbers = (Choices for hundreds) × (Choices for tens) × (Choices for units)
Total even numbers = 6 × 6 × 3 = 108
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 108
Quantity B = 108
The two quantities are equal.
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