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A company that ships boxes to a total of \(\mathrm{12}\) distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter.)
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:
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
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:
For example, with 3 colors (Red, Blue, Green):
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:
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: \(\mathrm{n + n(n-1)/2 = n(1 + (n-1)/2) = n(2 + n - 1)/2 = n(n+1)/2}\)
Now we need to find the smallest n where \(\mathrm{n(n+1)/2 ≥ 12}\).
Let's test small values:
n = 3:
n = 4:
n = 5:
Let's verify using our simplified formula: \(\mathrm{n(n+1)/2}\)
Process Skill: APPLY CONSTRAINTS - Testing systematically until we meet the minimum requirement
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.
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.
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.
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.