Arrange 2, 1, and 3: Formulating a Number Divisible by 2

To solve this problem, let's explore how to form a number divisible by 2 using the digits 2, 1, and 3. A number is divisible by 2 if the last digit is an even number, which, from our given digits, is only the digit 2.

Let's examine all possible permutations of these digits and identify which numbers are divisible by 2:

• Permutation 1: 213, where the last digit is 3 (not divisible by 2).
• Permutation 2: 231, where the last digit is 1 (not divisible by 2).
• Permutation 3: 123, where the last digit is 3 (not divisible by 2).
• Permutation 4: 132, where the last digit is 2 (divisible by 2).
• Permutation 5: 312, where the last digit is 2 (divisible by 2).
• Permutation 6: 321, where the last digit is 1 (not divisible by 2).

From our examination, the numbers 132 and 312 both end in 2, making them divisible by 2. Therefore, the correct choice according to the problem's options is to select both permutations that satisfy the condition.

Therefore, the solution to the problem is Answer b and c.