Question

How many 3-digit numbers can be formed using three distinct single digit prime numbers?

• A. 64
• B. 24
• C. 12
• D. 4

Hint:

First list all single-digit prime numbers, then count the ways to arrange three distinct ones in order.

Answer 1

The Correct Option is B

Step 1: List the single digit prime numbers.
The single digit numbers are \(0\) to \(9\). Among these, the prime numbers are \(2, 3, 5, 7\). So there are exactly \(4\) single digit primes available to choose from.

Step 2: Understand what is being asked.
We need 3-digit numbers where all three digits are distinct and each digit is one of \(2, 3, 5, 7\). Since none of these primes is \(0\), there is no risk of a number starting with \(0\), so every arrangement of three of these digits gives a valid 3-digit number.

Step 3: Count the arrangements.
We must choose 3 digits out of the 4 available primes and arrange them in order, since a 3-digit number cares about digit position (for example \(235\) and \(325\) are different numbers). This is a permutation problem. The number of ways to arrange \(3\) out of \(4\) distinct items is
\[ {}^4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24 \]
Here, there are \(4\) choices for the first digit, then \(3\) remaining choices for the second digit (since digits must be distinct), then \(2\) remaining choices for the third digit.

Step 4: Final Answer.
Multiplying these choices gives \(4 \times 3 \times 2 = 24\) distinct 3-digit numbers. \[ \boxed{24} \]