Question

• A. The quantity on the left is greater
• B. The quantity on the right is greater
• C. Both are equal
• D. The relationship cannot be determined without further information

Hint:

When forming numbers with constraints (like being even or odd), always fill the constrained position first (the units place in this case). This helps determine the remaining choices for the other positions correctly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves forming numbers with specific properties (even, two-digit, three-digit) from a given set of digits (prime numbers) without repetition. This is a permutation problem with constraints.
Step 2: Key Formula or Approach:
We will use the fundamental principle of counting. We identify the choices for each digit's place, starting with the most restricted position.
Step 3: Detailed Explanation:
For Column A:
First, identify the prime numbers less than 10. They are \{2, 3, 5, 7\}.
We need to form a three-digit even number without repetition using these digits.
For a number to be even, its units digit must be even. From our set \{2, 3, 5, 7\}, the only even digit is 2.
- Units place: Must be 2. So, there is only 1 choice.
- After placing 2 in the units place, the remaining digits are \{3, 5, 7\}.
- Hundreds place: We can choose any of the 3 remaining digits. So, there are 3 choices.
- Tens place: After filling the hundreds place, we have 2 digits left. So, there are 2 choices.
Total number of three-digit even numbers = (Choices for hundreds) \( \times \) (Choices for tens) \( \times \) (Choices for units)
\[ \text{Total numbers} = 3 \times 2 \times 1 = 6 \] So, Quantity A is 6.
For Column B:
First, identify the prime numbers less than 7. They are \{2, 3, 5\}.
We need to form a two-digit number without repetition using these digits.
This is an arrangement of 2 digits chosen from the 3 available digits.
- Tens place: We can choose any of the 3 digits. So, there are 3 choices.
- Units place: After filling the tens place, we have 2 digits left. So, there are 2 choices.
Total number of two-digit numbers = (Choices for tens) \( \times \) (Choices for units)
\[ \text{Total numbers} = 3 \times 2 = 6 \] This is also a permutation \(P(3, 2) = \frac{3!}{(3-2)!} = 6\).
So, Quantity B is 6.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 6
Quantity B = 6
The two quantities are equal.