50 students expressed their desire to participate in 3 types of swimming in the citys student championship. these types are: |-kroll, ||-brass, |||-butterfly. some students participated in one of the types, some in two, some in all tree: its known that16-participated in type |, 18-in ||, 24-in |||, 4-in |and ||, 5- in|| and |||, 8-in | and |||, 3- in all. find the number of students who participated in type |

We can use the principle of inclusion and exclusion to solve this problem.

Let:
- \( |A| \) represent the number of students who participated in type I (kroll).
- \( |B| \) represent the number of students who participated in type II (brass).
- \( |C| \) represent the number of students who participated in type III (butterfly).

The information given is as follows:
- \( |A| = 16 \) (students participated in type I),
- \( |B| = 18 \) (students participated in type II),
- \( |C| = 24 \) (students participated in type III),
- \( |A \cap B| = 4 \) (students participated in both type I and type II),
- \( |B \cap C| = 5 \) (students participated in both type II and type III),
- \( |A \cap C| = 8 \) (students participated in both type I and type III),
- \( |A \cap B \cap C| = 3 \) (students participated in all three types).

We are asked to find how many students participated in type I.

We start by using the inclusion-exclusion principle:

\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|
\]

Substitute the known values:

\[
|A \cup B \cup C| = 16 + 18 + 24 - 4 - 5 - 8 + 3 = 44
\]

This gives us that 44 students participated in at least one of the swimming types.

Since the total number of students is 50, the number of students who did not participate in any swimming type is:

\[
50 - 44 = 6
\]

Now, to find the number of students who participated **only** in type I, we can use the following formula:

\[
\text{Students only in } A = |A| - (|A \cap B| + |A \cap C| - |A \cap B \cap C|)
\]

Substitute the known values:

\[
\text{Students only in } A = 16 - (4 + 8 - 3) = 16 - 9 = 7
\]

Thus, **7 students participated only in type I (kroll)**.

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### Do you have any questions or need further details? Here are 5 related questions to expand on this topic:

1. How can the principle of inclusion and exclusion be applied in other types of problems?
2. What would happen if we changed the number of students participating in all three events?
3. Can we calculate how many students participated in exactly two types of swimming?
4. How many students participated only in type II (brass)?
5. How can we represent this situation using a Venn diagram?

**Tip:** When solving set problems, it's helpful to break them into smaller groups, such as "only in one set" or "in exactly two sets" to simplify calculations.

50 students expressed their desire to participate in 3 types of swimming in the city's student championship. These types are: |-kroll, ||-brass, |||-butterfly. Some students participated in one of the types, some in two, some in all three. It's known that 16 participated in type |, 18 in ||, 24 in |||, 4 in | and ||, 5 in || and |||, 8 in | and |||, 3 in all. Find the number of students who participated in type |.

Solve a set theory problem using the principle of inclusion and exclusion. Learn how 50 students participated in various swimming events—Kroll, Brass, and Butterfly—and how to find the number of students involved in each event. This example is ideal for students in Grades 9-11, covering key concepts in set theory and arithmetic.

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