If X and Y are sets of integers, X@Y denotes the set of integers that belong to set X or...

1. Translate the problem requirements: The symbol \(\mathrm{X@Y}\) represents a "symmetric difference" - elements that are in X OR Y, but NOT in both sets. We need to count how many integers are in this combined set.
2. Visualize the set relationships: Picture the overlap between sets X and Y to understand which elements belong to \(\mathrm{X@Y}\) and which don't.
3. Identify elements to exclude: Since \(\mathrm{X@Y}\) excludes elements that are in BOTH sets, we need to remove the intersection from our count.
4. Calculate using basic arithmetic: Add elements unique to X plus elements unique to Y to get the total count for \(\mathrm{X@Y}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to find. The symbol \(\mathrm{X@Y}\) represents something called a "symmetric difference" - but don't worry about fancy terms. Think of it this way: \(\mathrm{X@Y}\) includes all the integers that are in ONLY one of the two sets, but NOT in both sets at the same time.

So if an integer appears in set X but not in set Y, it goes into \(\mathrm{X@Y}\). If an integer appears in set Y but not in set X, it also goes into \(\mathrm{X@Y}\). But if an integer appears in BOTH X and Y, it gets excluded from \(\mathrm{X@Y}\).

Our job is to count how many integers end up in this \(\mathrm{X@Y}\) set.

Process Skill: TRANSLATE - Converting the symbol notation into plain English understanding

2. Visualize the set relationships

Let's picture this with a simple example first. Imagine:
- Set X contains: \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)
- Set Y contains some numbers that overlap with X, and some that don't
- 6 numbers appear in BOTH sets

Now, think of it like two circles that overlap. The overlapping part contains the 6 integers that are in both X and Y. The non-overlapping parts contain integers that are in only one set or the other.

\(\mathrm{X@Y}\) will include everything in the non-overlapping parts, but EXCLUDE everything in the overlapping part.

Process Skill: VISUALIZE - Using concrete mental images to understand abstract set operations

3. Identify elements to exclude

Here's the key insight: since \(\mathrm{X@Y}\) excludes elements that are in BOTH sets, we need to be careful not to count the overlapping elements.

From our given information:
- 6 integers appear in both X and Y (this is the overlap)
- These 6 integers will NOT be part of \(\mathrm{X@Y}\)

So we need to find how many integers are in X but not in Y, plus how many integers are in Y but not in X.

4. Calculate using basic arithmetic

Let's count step by step:

Step 1: Find integers that are in X but NOT in Y
- Set X has 10 integers total
- 6 of these are also in Y
- So integers in X but not in Y = 10 - 6 = 4 integers

Step 2: Find integers that are in Y but NOT in X
- Set Y has 18 integers total
- 6 of these are also in X
- So integers in Y but not in X = 18 - 6 = 12 integers

Step 3: Add them together for \(\mathrm{X@Y}\)
- \(\mathrm{X@Y}\) = (integers only in X) + (integers only in Y)
- \(\mathrm{X@Y}\) = 4 + 12 = 16 integers

4. Final Answer

\(\mathrm{X@Y}\) consists of 16 integers.

Looking at our answer choices, this matches option (B) 16.

We can verify this makes sense: We started with 10 + 18 = 28 total integers if we counted everything. But since 6 integers are counted twice (they appear in both sets), and \(\mathrm{X@Y}\) specifically excludes these overlapping integers, we get 28 - 6 - 6 = 16 integers in \(\mathrm{X@Y}\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the \(\mathrm{X@Y}\) operation
Students often confuse \(\mathrm{X@Y}\) with regular set union (\(\mathrm{X} \cup \mathrm{Y}\)) and think it means "all elements in either X or Y." They miss the crucial "but not both" part of the definition. This leads them to incorrectly include the overlapping elements instead of excluding them.

2. Misinterpreting "6 integers are in both X and Y"
Some students may think this means there are 6 additional integers somewhere, rather than understanding that these 6 integers are already counted within the 10 integers of set X and within the 18 integers of set Y. This misunderstanding leads to incorrect counting strategies.

Errors while executing the approach

1. Double-counting or incorrect subtraction
Even with the right approach, students often make arithmetic errors when calculating "integers only in X" and "integers only in Y." They might forget to subtract the overlapping 6 integers from both sets, or they might subtract incorrectly (like doing 10 - 6 - 6 instead of 10 - 6).

2. Using the wrong formula
Students sometimes try to use the inclusion-exclusion principle incorrectly, calculating \(|\mathrm{X}| + |\mathrm{Y}| - |\mathrm{X} \cap \mathrm{Y}|\) = 10 + 18 - 6 = 22, which gives the size of \(\mathrm{X} \cup \mathrm{Y}\) (regular union), not \(\mathrm{X@Y}\) (symmetric difference). They forget that symmetric difference requires subtracting the intersection twice.

Errors while selecting the answer

1. Choosing the union result instead of symmetric difference
After getting confused during execution, students often end up with 22 (which is \(|\mathrm{X} \cup \mathrm{Y}|\)) and select answer choice (C) instead of the correct answer (B) 16. This happens because 22 "feels right" since it's close to the sum of the set sizes, making it a tempting wrong answer.