This is a direct counting problem, best solved by filling the three digit slots of the number one at a time, from left to right, using the slot-filling (multiplication) method instead of the permutation formula directly.
- Find the pool of digits: Among the digits $0$ through $9$, the prime numbers are $2, 3, 5, 7$. That gives a pool of exactly $4$ digits to pick from, and since $0$ is not prime, every number built from these digits is automatically a valid 3-digit number with no leading-zero problem.
- Fill the hundreds place: Any of the $4$ primes can go here, so there are $4$ options.
- Fill the tens place: The digits must be distinct, so one prime has already been used in the hundreds place. That leaves $3$ primes for the tens place.
- Fill the units place: Two primes are now used up, leaving $2$ primes for the units place.
Multiplying the choices for each slot gives the total count of distinct numbers: $4 \times 3 \times 2 = 24$.
Let's summarize:
- There are only $4$ single digit primes: $2, 3, 5, 7$.
- Filling three slots one at a time with shrinking choices ($4$, then $3$, then $2$) gives $24$ total 3-digit numbers.
So $24$ distinct 3-digit numbers can be formed, matching option (B).