A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If...

1. Translate the problem requirements: We need to find the minimum number of colors such that we can create at least 12 unique identifiers, where each identifier is either one single color OR one pair of two different colors (order doesn't matter in pairs).
2. Set up the counting framework: Determine how many unique identifiers we can create with n colors - this includes counting single colors and counting pairs of different colors.
3. Apply combination counting logic: For n colors, we get n single-color identifiers plus \(\mathrm{C(n,2)}\) pair identifiers, giving us total identifiers = \(\mathrm{n + n(n-1)/2}\).
4. Test values systematically: Start with small values of n and calculate total identifiers until we reach or exceed 12, then identify the minimum n that works.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have 12 distribution centers, and each one needs a unique identifier using colors. Think of this like creating unique name tags for 12 different locations.

The rules are:

• Each center gets either ONE single color (like "Red") OR a pair of two different colors (like "Red-Blue")
• Every center must have a different identifier
• When using pairs, order doesn't matter (so "Red-Blue" is the same as "Blue-Red")

Our goal is to find the smallest number of colors that can create at least 12 unique identifiers.

Process Skill: TRANSLATE - Converting the problem language into clear counting requirements

2. Set up the counting framework

Now let's think about how many unique identifiers we can make with any number of colors. Let's call the number of colors "n".

If we have n colors, we can create identifiers in two ways:

Single-color identifiers: Each color by itself makes one identifier. So with n colors, we get exactly n single-color identifiers.

Pair identifiers: We need to count how many different pairs we can make from n colors. Let's think about this step by step:

• If I pick the first color, I can pair it with any of the remaining (n-1) colors
• But since order doesn't matter, I need to avoid double-counting

For example, with 3 colors (Red, Blue, Green):

• Single identifiers: Red, Blue, Green (that's 3)
• Pair identifiers: Red-Blue, Red-Green, Blue-Green (that's 3 pairs)
• Total: 3 + 3 = 6 unique identifiers

3. Apply combination counting logic

Let me show you the pattern for counting pairs more systematically:

With n colors, the number of different pairs is what mathematicians call "n choose 2," but let's build this from scratch:

• First color can pair with (n-1) others
• Second color can pair with (n-2) others (we already counted its pair with the first)
• Third color can pair with (n-3) others
• And so on...

This gives us: \(\mathrm{(n-1) + (n-2) + (n-3) + ... + 1 = n(n-1)/2}\)

So our total number of unique identifiers with n colors is:
Total identifiers = \(\mathrm{n + n(n-1)/2}\)

We can simplify this: \(\mathrm{n + n(n-1)/2 = n(1 + (n-1)/2) = n(2 + n - 1)/2 = n(n+1)/2}\)

4. Test values systematically

Now we need to find the smallest n where \(\mathrm{n(n+1)/2 ≥ 12}\).

Let's test small values:

n = 3:

• Single identifiers: 3
• Pair identifiers: \(\mathrm{3(2)/2 = 3}\)
• Total: 3 + 3 = 6 identifiers ❌ (need 12)

n = 4:

• Single identifiers: 4
• Pair identifiers: \(\mathrm{4(3)/2 = 6}\)
• Total: 4 + 6 = 10 identifiers ❌ (need 12)

n = 5:

• Single identifiers: 5
• Pair identifiers: \(\mathrm{5(4)/2 = 10}\)
• Total: 5 + 10 = 15 identifiers ✓ (exceeds 12)

Let's verify using our simplified formula: \(\mathrm{n(n+1)/2}\)

• For n = 4: \(\mathrm{4(5)/2 = 10}\) ❌
• For n = 5: \(\mathrm{5(6)/2 = 15}\) ✓

Process Skill: APPLY CONSTRAINTS - Testing systematically until we meet the minimum requirement

4. Final Answer

With 5 colors, we can create exactly 15 unique identifiers (5 single-color + 10 pairs), which is more than the required 12. With 4 colors, we can only create 10 identifiers, which is insufficient.

Therefore, the minimum number of colors needed is 5.

The answer is B. 5.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint about order not mattering for pairs
Students often misread "order of colors in a pair does not matter" and think they need to count ordered pairs (like Red-Blue AND Blue-Red as different identifiers). This leads them to use the formula for permutations instead of combinations, doubling their pair count and arriving at an incorrect minimum number of colors.

2. Forgetting that single colors are also valid identifiers
Many students focus only on counting pairs and forget that each individual color can also serve as a unique identifier by itself. This oversight leads them to only count combinations of 2 colors from n colors, missing the additional n single-color identifiers, resulting in an undercount of total possible identifiers.

3. Setting up an equality instead of inequality constraint
Students sometimes look for exactly 12 identifiers rather than "at least 12" identifiers. This misinterpretation can lead them to more complex calculations or incorrect reasoning about what constitutes the "minimum" number of colors needed.

Errors while executing the approach

1. Arithmetic errors in combination formula
When calculating \(\mathrm{n(n-1)/2}\) for pairs, students frequently make basic arithmetic mistakes, especially with fractions. For example, when calculating \(\mathrm{5(4)/2}\), they might get 9 instead of 10, or when calculating \(\mathrm{4(3)/2}\), they might get 5 instead of 6.

2. Incorrect application of the combination formula
Students may confuse the combinations formula and use \(\mathrm{n!/(2!(n-2)!)}\) incorrectly, or mix up which values to substitute for n. Some might also forget to add the single-color identifiers to their pair count, calculating only the pairs.

3. Stopping testing too early
After finding that n=4 gives 10 identifiers (which is less than 12), some students might jump to conclusions without systematically checking n=5, or they might incorrectly conclude that they need to double the number of colors rather than incrementally testing the next value.

Errors while selecting the answer

1. Selecting the total number of identifiers instead of number of colors
After correctly calculating that 5 colors give 15 total identifiers, students sometimes mistakenly select 15 as their answer instead of 5 (the number of colors). They confuse what the question is asking for - the minimum number of colors, not the total number of possible identifiers.