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:

The single digit primes are 2, 3, 5, 7. Since the three digits must be distinct, count ordered selections: 4 x 3 x 2.

Answer 1

The Correct Option is B

Step 1: List the single digit prime numbers.
A single digit number runs from 0 to 9. The prime numbers among these are 2, 3, 5, and 7, so there are exactly 4 single digit primes to choose from. None of them is 0, so there is no risk of a leading zero in any 3-digit number we build.

Step 2: Understand the condition "three distinct" primes.
The 3-digit number must use three different primes from this list of 4, with no digit repeated. So we are really asking: in how many ways can we pick 3 of these 4 primes and place them in order to form a number, where changing the order gives a different number (for example, 235 and 253 are different 3-digit numbers).

Step 3: Count using the multiplication principle.
Fill the hundreds place first: any of the 4 primes can go there, giving 4 choices. Once the hundreds digit is fixed, the tens place must be a different prime, so only 3 primes remain, giving 3 choices. Once both of those digits are fixed, the units place must be different from both, so only 2 primes remain, giving 2 choices. Multiplying the choices for each position:
\[ 4 \times 3 \times 2 = 24 \]

Step 4: Rule out the other options.
64 equals \(4^3\), which would be the count if digits were allowed to repeat, but the question requires distinct digits, so 64 is wrong. 12 and 4 are both smaller than the correct count of arrangements and undercount the actual number of orderings possible.

Final Answer:
The number of 3-digit numbers using three distinct single digit primes is 24. \[ \boxed{24} \]