Question

How many 4-digit numbers (between 1000 and 9999) can be formed using the digits \(2, 3, 5, 7, 8\) such that:

• Each number contains both \(3\) and \(7\) exactly once,
• No digit is repeated.

• A. 24
• B. 36
• C. 28
• D. 72

Hint:

When forming numbers with specific digits and no repetition, consider restrictions on the first digit carefully, especially to avoid leading zero or disallowed digits.

Answer 1

The Correct Option is D

Step 1: Choose the other two digits
Since \(3\) and \(7\) must both appear exactly once, we select the other two digits from \(\{2, 5, 8\}\). Number of ways to choose these two digits: \[ \binom{3}{2} = 3 \] Step 2: Arrange the four chosen digits
Each chosen set of 4 digits (which always includes 3 and 7 plus the chosen two) can be permuted in: \[ 4! = 24 \] ways.
Step 3: Check leading digit restrictions
Since zero is not in the digit set, there is no restriction on the leading digit.
Step 4: Calculate total number of 4-digit numbers
\[ 3 \times 24 = 72 \] Final answer: \[ \boxed{72} \]