For a certain cube, 3 different edges that do not all meet at a single vertex are to be painted...

1. Translate the problem requirements: We need to select 3 different edges from a cube's 12 edges, but with the restriction that these 3 edges cannot all meet at a single vertex (since each vertex has exactly 3 edges meeting at it).
2. Calculate total unrestricted selections: Find the total number of ways to choose any 3 edges from the 12 edges of a cube without any restrictions.
3. Identify and count restricted cases: Determine how many ways we can select 3 edges that DO all meet at a single vertex (the cases we need to exclude).
4. Apply subtraction principle: Subtract the restricted cases from the total unrestricted cases to get our final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're working with. A cube has 12 edges total - think of a dice or a box. Each corner (vertex) of the cube has exactly 3 edges meeting at it, and a cube has 8 corners total.

We need to select 3 different edges to paint red, but there's an important restriction: these 3 edges cannot all meet at the same corner. Why is this restriction important? Because if all 3 edges meet at one corner, that corner would have all its edges painted red, which apparently isn't what we want.

So our task becomes: count all possible ways to choose 3 edges, then subtract the "bad" cases where all 3 edges meet at one vertex.

Process Skill: TRANSLATE - Converting the geometric constraint into a clear mathematical counting problem

2. Calculate total unrestricted selections

First, let's find how many ways we can choose any 3 edges from the 12 edges, ignoring the restriction for now.

This is like choosing 3 items from a collection of 12 items, where order doesn't matter. In everyday terms, if you had 12 different colored pencils and wanted to pick 3 of them, how many different combinations could you make?

The mathematical way to express this is "12 choose 3" or \(\mathrm{C(12,3)}\).

Calculating step by step: \(\mathrm{C(12,3)} = \frac{12!}{3! \times 9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1,320}{6} = 220\)

So there are 220 total ways to select any 3 edges from the cube.

3. Identify and count restricted cases

Now we need to count the "bad" cases - selections where all 3 edges meet at a single vertex.

Here's the key insight: each vertex of a cube has exactly 3 edges meeting at it. So if we want all 3 of our selected edges to meet at one vertex, we must choose ALL 3 edges that meet at that vertex.

Since a cube has 8 vertices, and each vertex gives us exactly one way to choose 3 edges that all meet there, we have exactly 8 "bad" cases to exclude.

Think of it this way: for each of the 8 corners of the cube, there's exactly one way to pick the 3 edges that meet at that corner.

Process Skill: VISUALIZE and APPLY CONSTRAINTS - Understanding the geometric structure to count restricted cases accurately

4. Apply subtraction principle

Now we can find our answer using the subtraction principle:

Total valid selections = Total unrestricted selections - Restricted cases Total valid selections = 220 - 8 = 212

Let's verify this makes sense: we started with 220 ways to choose any 3 edges, then removed the 8 cases where all 3 edges meet at a single vertex, leaving us with 212 valid ways to select 3 edges that don't all meet at one vertex.

Final Answer

The number of different selections of 3 different edges that do not all meet at a single vertex is 212.

Looking at our answer choices, this matches choice D: 212.

The answer is D.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "do not all meet at a single vertex"

Students often misread this as "no two edges can share a vertex" or "all edges must be completely separate." This leads them to think the problem is much more restrictive than it actually is. The constraint only prohibits having all THREE edges meeting at the SAME vertex - it's perfectly fine for some edges to share vertices as long as all three don't converge at one point.

2. Attempting direct counting instead of using complementary counting

Many students try to directly count valid configurations by considering different cases (like "2 edges share a vertex, 1 doesn't" or "no edges share vertices"). This approach becomes extremely complex and error-prone because there are many overlapping subcases to track. The complementary approach (total cases minus restricted cases) is much cleaner but requires recognizing this strategic insight.

3. Incorrectly identifying the structure of a cube

Some students may forget that a cube has exactly 12 edges and 8 vertices, or may not realize that exactly 3 edges meet at each vertex. Without this foundational knowledge, they cannot set up the counting problem correctly from the start.

Errors while executing the approach

1. Calculation errors in computing \(\mathrm{C(12,3)}\)

Students frequently make arithmetic mistakes when calculating combinations, especially in the step \(\mathrm{C(12,3)} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1}\). Common errors include forgetting to divide by 3! or making multiplication/division mistakes that lead to incorrect values like 240 instead of 220.

2. Miscounting the number of restricted cases

Even when students understand that each vertex creates one restricted case, they might miscount the number of vertices (thinking a cube has 6 vertices like faces, instead of 8) or double-count cases, leading to subtracting the wrong number from the total.

Errors while selecting the answer

1. Selecting 220 instead of 212

Students who correctly calculate the total number of ways to choose 3 edges (220) but forget to subtract the restricted cases will select an answer choice closest to 220. Looking at the options, they might be tempted by other choices, but the key error is stopping after finding the unrestricted total without applying the constraint.