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

This is a direct application of the permutation formula, and it helps to first be clear on which digits are even in play.

1. Identify the digit pool: single digit prime numbers are 2, 3, 5, 7, giving a pool of 4 digits, none of which is zero.
2. Decide the counting rule: since the three digits used in each number must be distinct, and order matters (237 and 273 are different numbers), this calls for permutations, written $P(n,r)$ for choosing and arranging $r$ items out of $n$.
3. Apply the formula: here $n=4$ and $r=3$, so $P(4,3) = \dfrac{4!}{(4-3)!} = \dfrac{4!}{1!} = \dfrac{24}{1} = 24$.
4. Sanity check by direct multiplication: the hundreds digit has 4 choices, the tens digit has 3 remaining choices, and the units digit has 2 remaining choices, giving $4 \times 3 \times 2 = 24$, which matches the formula.

Both the formula and the direct slot-by-slot count agree, so the number of valid 3-digit numbers is 24.

Let's summarize:

• There are 4 single digit primes: 2, 3, 5, 7.
• Choosing 3 distinct ones and arranging them gives $P(4,3)=24$.

The correct count is 24, option (B).