A 5-digit identification code is to be created as a sequence that contains each integer in the set {1, 2,...

1. Translate the problem requirements: We need to arrange digits \(\{1, 2, 3, 4, 5\}\) in a 5-digit sequence where each digit appears exactly once, but with the restriction that even numbers (2 and 4) cannot be placed next to each other.
2. Identify the constraint structure: Recognize that we have 3 odd numbers \(\{1, 3, 5\}\) and 2 even numbers \(\{2, 4\}\), where the even numbers must be separated by at least one odd number.
3. Determine valid placement patterns: Find all possible ways to position the two even numbers in a 5-position sequence such that they are not adjacent.
4. Calculate arrangements for each pattern: For each valid placement pattern of even numbers, count how many ways we can arrange the specific even numbers in those positions and the odd numbers in the remaining positions.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to do in everyday language. We have five digits: \(\{1, 2, 3, 4, 5\}\) and we need to arrange them in a row to create a 5-digit code. Each digit must be used exactly once - so this is like arranging 5 different cards in a line.

The key restriction is that even numbers (2 and 4) cannot be placed next to each other. Think of it like two people who don't get along - they can't sit next to each other at a dinner table.

So we're looking for arrangements like: 1-2-3-4-5, but NOT arrangements like: 1-2-4-3-5 (because 2 and 4 are next to each other).

Process Skill: TRANSLATE - Converting the constraint "even integers next to each other" into concrete understanding of what arrangements are forbidden.

2. Identify the constraint structure

Let's organize our numbers by type:

• Odd numbers: \(\{1, 3, 5\}\) - we have 3 of these
• Even numbers: \(\{2, 4\}\) - we have 2 of these

The constraint means our two even numbers (2 and 4) must be separated by at least one odd number. Think of the odd numbers as "buffers" that keep the even numbers apart.

Since we have 3 odd numbers and only 2 even numbers that need to be separated, we should have enough "buffers" to make this work in multiple ways.

3. Determine valid placement patterns

Now let's figure out where we can place our two even numbers in a 5-position sequence so they're not next to each other.

Let's call the positions 1, 2, 3, 4, 5. We need to choose 2 of these 5 positions for our even numbers, but they can't be adjacent.

Let's list all possible ways to choose 2 non-adjacent positions:

• Positions (1,3): Even numbers in slots 1 and 3
• Positions (1,4): Even numbers in slots 1 and 4
• Positions (1,5): Even numbers in slots 1 and 5
• Positions (2,4): Even numbers in slots 2 and 4
• Positions (2,5): Even numbers in slots 2 and 5
• Positions (3,5): Even numbers in slots 3 and 5

We can verify: (1,2), (2,3), (3,4), and (4,5) are NOT allowed because these positions are adjacent.

So we have exactly 6 valid placement patterns.

Process Skill: CONSIDER ALL CASES - Systematically listing all possible non-adjacent position pairs to ensure we don't miss any valid arrangements.

4. Calculate arrangements for each pattern

For each of our 6 valid placement patterns, we need to count the total arrangements:

For any single pattern:

• We can arrange the 2 even numbers \(\{2, 4\}\) in the chosen even positions in \(2! = 2 \times 1 = 2\) ways
• We can arrange the 3 odd numbers \(\{1, 3, 5\}\) in the remaining 3 positions in \(3! = 3 \times 2 \times 1 = 6\) ways

So for each pattern: \(2 \times 6 = 12\) total arrangements

Since we have 6 valid patterns: \(6 \times 12 = 72\) total arrangements

Let's verify with a concrete example using pattern (1,3):

• Even positions 1,3 can be filled as: (2,4) or (4,2)
• Remaining positions 2,4,5 filled with \(\{1,3,5\}\) can be done in 6 ways: (1,3,5), (1,5,3), (3,1,5), (3,5,1), (5,1,3), (5,3,1)
• This gives us \(2 \times 6 = 12\) arrangements for just this one pattern

4. Final Answer

The total number of valid identification codes is 72.

This matches answer choice C.

To double-check: If there were no restrictions, we'd have \(5! = 120\) total arrangements. Our answer of 72 is less than 120, which makes sense since we're excluding arrangements where even numbers are adjacent.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint

Students often misread "even integers next to each other" as "odd integers next to each other" or think it means "no even integers can be used." This fundamental misunderstanding leads them down completely wrong solution paths. They need to clearly identify that only 2 and 4 cannot be adjacent, while all other adjacencies are allowed.

2. Attempting to use total minus forbidden approach without proper setup

Many students try to calculate \(5! -\) (arrangements with 2 and 4 adjacent) but struggle with correctly counting the forbidden arrangements. This approach requires treating "24" or "42" as single units, which adds complexity. The direct counting method shown in the solution is more straightforward and less error-prone.

3. Overlooking the systematic case-by-case structure

Students may recognize they need to place even numbers in non-adjacent positions but fail to systematically list all possible position pairs. They might randomly start arranging without a clear framework, leading to either missing valid cases or double-counting arrangements.

Errors while executing the approach

1. Miscounting non-adjacent position pairs

When listing position pairs like (1,3), (1,4), (1,5), (2,4), (2,5), (3,5), students frequently include adjacent pairs like (1,2) or (3,4) in their count. They may also miss valid pairs like (2,5), resulting in an incorrect base count that throws off their entire calculation.

2. Arithmetic errors in factorial calculations

Students often make basic computational mistakes when calculating \(2! \times 3! = 2 \times 6 = 12\) for each pattern, or when multiplying \(6\) patterns \(\times 12\) arrangements \(= 72\). These arithmetic errors are especially common under time pressure, leading to selecting incorrect answer choices.

3. Confusing which numbers go in which positions

Even with the correct approach, students sometimes mix up whether they're arranging the 2 even numbers in the chosen positions or the 3 odd numbers in the remaining positions. This confusion can lead them to calculate \(3! \times 2!\) instead of \(2! \times 3!\), though this particular error doesn't affect the final answer since both equal 12.

Errors while selecting the answer

No likely faltering points - once students correctly execute the calculation and arrive at 72, this number directly matches answer choice C, making selection straightforward.