The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x, y}, {w, x,...

1. Translate the problem requirements: We need to find how many subsets of the 4-element set \(\{\mathrm{w, x, y, z}\}\) contain the element w. A subset "contains w" means w must be one of the elements in that subset.
2. Identify the constraint and remaining choices: Since w must be included in every subset we count, we need to determine what choices we have for the remaining elements x, y, and z.
3. Apply the choice principle for remaining elements: For each of the remaining elements (x, y, z), we have exactly two choices - either include it in the subset or don't include it. Count the total combinations possible.
4. Verify with systematic enumeration: List out all possible subsets containing w to confirm our count and ensure we haven't missed any or double-counted.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in plain English. We have a set \(\{\mathrm{w, x, y, z}\}\) with 4 elements, and we want to count how many subsets contain the element w.

What does "contain w" mean? It simply means that w must be one of the elements listed in the subset. For example, \(\{\mathrm{w, x}\}\) contains w, but \(\{\mathrm{x, y}\}\) does not contain w.

So we're counting subsets where w is definitely included, and we need to figure out what to do with the remaining elements x, y, and z.

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

2. Identify the constraint and remaining choices

Since w must be included in every subset we count, we can think of this problem differently. Instead of asking "which subsets contain w?", we can ask "if w is already chosen, what are our options for the remaining elements?"

The remaining elements are x, y, and z. These are the elements we have flexibility with - we can either include them or leave them out of our subset.

This transforms our problem into: "In how many ways can we form subsets from the remaining elements \(\{\mathrm{x, y, z}\}\) and then add w to each one?"

3. Apply the choice principle for remaining elements

Now let's think about this step by step. For each of the remaining elements x, y, and z, we have exactly two choices:

- Include it in the subset, OR
- Don't include it in the subset

For element x: 2 choices (include or exclude)
For element y: 2 choices (include or exclude)
For element z: 2 choices (include or exclude)

Since these choices are independent of each other, we multiply the number of choices:
Total combinations = \(\mathrm{2 \times 2 \times 2 = 8}\)

Each of these 8 combinations gives us a different subset that contains w.

4. Verify with systematic enumeration

Let's list out all possible subsets containing w to double-check our answer:

1. \(\{\mathrm{w}\}\) - we include w, exclude x, y, z
2. \(\{\mathrm{w, x}\}\) - we include w and x, exclude y, z
3. \(\{\mathrm{w, y}\}\) - we include w and y, exclude x, z
4. \(\{\mathrm{w, z}\}\) - we include w and z, exclude x, y
5. \(\{\mathrm{w, x, y}\}\) - we include w, x, y, exclude z
6. \(\{\mathrm{w, x, z}\}\) - we include w, x, z, exclude y
7. \(\{\mathrm{w, y, z}\}\) - we include w, y, z, exclude x
8. \(\{\mathrm{w, x, y, z}\}\) - we include all elements

Counting these up: 1, 2, 3, 4, 5, 6, 7, 8 - exactly 8 subsets!

This confirms our calculation using the choice principle.

Final Answer

The number of subsets of \(\{\mathrm{w, x, y, z}\}\) that contain w is 8.

Looking at our answer choices, this corresponds to choice D) Eight.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint

Students often confuse "subsets that contain w" with "subsets that contain ONLY w" or "subsets that start with w". The key insight is that w must be included, but other elements can be freely chosen. This misinterpretation leads to undercounting the subsets.

2. Attempting to use the total subset formula incorrectly

Some students know that a set with n elements has \(\mathrm{2^n}\) subsets total, so they calculate \(\mathrm{2^4 = 16}\) for the set \(\{\mathrm{w,x,y,z}\}\) and then try to divide by something or apply some ratio. This approach doesn't directly help with the constraint that w must be included.

3. Failing to recognize this as a conditional counting problem

Students may not realize they can fix w as "always included" and then count the choices for the remaining elements \(\{\mathrm{x,y,z}\}\). Instead, they might try to enumerate all possible subsets and then filter, which is more error-prone.

Errors while executing the approach

1. Arithmetic errors in the multiplication

When calculating \(\mathrm{2 \times 2 \times 2}\) for the three remaining elements, students might make simple multiplication errors, especially under time pressure, getting 6 or 4 instead of 8.

2. Missing subsets during enumeration

If students choose to list out all subsets containing w, they often miss one or two subsets, particularly the singleton \(\{\mathrm{w}\}\) or the complete set \(\{\mathrm{w,x,y,z}\}\). The systematic enumeration requires careful organization.

3. Double-counting or incorrect categorization

When listing subsets, students might accidentally count the same subset twice or incorrectly include subsets that don't contain w (like \(\{\mathrm{x,y,z}\}\)).

Errors while selecting the answer

1. Selecting the wrong answer choice despite correct calculation

Students might correctly calculate 8 subsets but then select choice C) Seven due to rushed reading or mismatching their work with the answer choices.

2. Confusing with the given example

The question mentions that \(\{\mathrm{w,x,y}\}\) has specific subsets listed. Students might incorrectly think the answer should be 7 (the number of non-empty subsets from the example) rather than focusing on their own calculation.