How many 3-digit numbers can be formed using three distinct single digit prime numbers?
Show Hint
- 64
- 24
- 12
- 4
The Correct Option is B
Solution and Explanation
A prime number has exactly two factors, 1 and itself. Among the single digits 0 to 9, the prime numbers are 2, 3, 5 and 7 (1 is not prime, and 0, 4, 6, 8, 9 each have more than two factors). So there are exactly 4 single digit primes to choose from.
Step 2: Understand what is being asked.
We need 3-digit numbers where all three digits are distinct and each digit is one of 2, 3, 5, 7. Since the digits are chosen from a set of 4 primes and must stay distinct within the number, and the order of the digits matters (for example 237 and 273 are different numbers), this is a permutation problem: choosing and arranging 3 primes out of 4 in order.
Step 3: Apply the permutation formula.
The number of ways to arrange 3 distinct items chosen from 4 distinct items, where order matters, is \(^4P_3 = \dfrac{4!}{(4-3)!} = \dfrac{24}{1} = 24\).
Step 4: Verify by direct counting.
For the hundreds digit, there are 4 choices (2, 3, 5 or 7). Since one prime has already been used and digits must stay distinct, the tens digit has 3 remaining choices, and the units digit then has 2 remaining choices. Multiplying these choices gives \(4 \times 3 \times 2 = 24\).
Step 5: Confirm none of the digits is zero, so there is no leading zero issue.
Since 2, 3, 5 and 7 are all nonzero, every one of these 24 arrangements is automatically a valid 3-digit number, because the hundreds digit is never 0.
Step 6: Compare with the options.
The count 24 matches option (B). Option (A) 64 would be the count if repetition of digits were allowed (\(4^3 = 64\)), which is not the case here since the primes must be distinct. Option (C) 12 and option (D) 4 are too small and do not correspond to a correct counting method for this problem.
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