Let the Universal Set, S, have 215 elements. A and B are subsets of S. Set A contains 70 elements and Set B contains 99 elements. If the total number of elements in either A or B is 168, how many elements are in A but not in B?

We are given the following information:

- The universal set \( S \) has 215 elements.
- Set \( A \) has 70 elements: \( |A| = 70 \).
- Set \( B \) has 99 elements: \( |B| = 99 \).
- The number of elements in either \( A \) or \( B \) is 168: \( |A \cup B| = 168 \).

We are asked to find the number of elements in \( A \) but not in \( B \), denoted by \( |A - B| \).

### Key Formula:
We can use the principle of inclusion and exclusion to solve this. The formula for the union of two sets is:
\[
|A \cup B| = |A| + |B| - |A \cap B|
\]
Substitute the known values into the formula:
\[
168 = 70 + 99 - |A \cap B|
\]
Simplifying this:
\[
168 = 169 - |A \cap B|
\]
\[
|A \cap B| = 169 - 168 = 1
\]

So, the number of elements in both \( A \) and \( B \) is \( |A \cap B| = 1 \).

### Now, to find \( |A - B| \):
The number of elements in \( A \) but not in \( B \) is:
\[
|A - B| = |A| - |A \cap B|
\]
Substitute the known values:
\[
|A - B| = 70 - 1 = 69
\]

Thus, the number of elements in \( A \) but not in \( B \) is \( \boxed{69} \).

Would you like more details or have any questions? Here are some related questions you can explore:

1. How many elements are in \( B - A \)?
2. How many elements are in \( A \cap B \)?
3. What is the number of elements in the complement of \( A \cup B \)?
4. If \( |S| \) were different, how would that affect \( A \) and \( B \)?
5. Can you generalize this problem for any two sets \( A \) and \( B \)?

**Tip:** Always verify if you need the intersection or complement of sets when dealing with set operations!

Learn how to solve a set theory problem where you are given a universal set of 215 elements, and sets A and B with 70 and 99 elements, respectively. Using the inclusion-exclusion principle, we find the number of elements in A but not in B. Suitable for Grades 9-12.

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