Step 1: Identify the pool of digits available. The single digit primes are 2, 3, 5 and 7 (0, 1, 4, 6, 8, 9 are not prime, since 1 is not considered prime and the rest have more than two factors). This gives a pool of exactly 4 digits.
Step 2: Fill the 3-digit number position by position. A 3-digit number has a hundreds place, a tens place and a units place. Since the digits must be distinct, once a digit is used in one place it cannot be reused in another place.
Step 3: Count the choices for each place. For the hundreds place, any of the 4 primes can be used, giving 4 options. Once that digit is fixed, only 3 primes remain for the tens place, giving 3 options. After the tens digit is also fixed, only 2 primes remain for the units place, giving 2 options.
Step 4: Multiply the choices using the fundamental counting principle. The total number of distinct 3-digit numbers equals the product of the choices at each place: $4 \times 3 \times 2 = 24$.
Step 5: Sanity check with a smaller list. Fix the hundreds digit as 2. The remaining two places can be filled with any 2 of $\{3, 5, 7\}$ in order, which gives $3 \times 2 = 6$ numbers (235, 237, 253, 257, 273, 275). Since there are 4 choices for the hundreds digit and each gives 6 numbers by the same logic, the total is $4 \times 6 = 24$, confirming the earlier count.
Step 6: Eliminate the wrong options. 64 would only be correct if digits could repeat ($4^3 = 64$), which the question does not allow since it explicitly asks for distinct primes. 12 and 4 undercount the actual number of valid arrangements.
