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:

There are only 4 single digit primes (2, 3, 5, 7). Pick 3 distinct ones and arrange them: this is a permutation, $P(4,3)$.

Answer 1

The Correct Option is B

Step 1: List the single digit prime numbers.
The single digit primes are 2, 3, 5, and 7. There are exactly 4 such digits.

Step 2: Understand what is being asked.
We need 3-digit numbers where all three digits are distinct and each digit is chosen from the set \(\{2,3,5,7\}\). Since digits must be distinct, this is a permutation problem, not a problem with repetition allowed.

Step 3: Count the choices for each position.
For the hundreds place, there are 4 possible digits to choose from.
Once that digit is used, 3 digits remain for the tens place.
Once two digits are used, 2 digits remain for the units place.
None of these digits is 0, so there is no need to worry about a leading zero.

Step 4: Multiply the choices.
\[ 4 \times 3 \times 2 = 24 \]
This is the same as computing the permutation \(P(4,3) = \dfrac{4!}{(4-3)!} = \dfrac{24}{1} = 24\).

Step 5: Rule out the other options.
64 would be the count if repetition were allowed (\(4^3=64\)), which is not the case here since digits must be distinct. 12 and 4 are both smaller than the correct permutation count and do not correspond to any consistent counting rule for this problem.

Final Answer:
24 distinct 3-digit numbers can be formed. \[ \boxed{24} \]