In a class of 60 students, 23 play hockey, 15 play basketball,20 play cricket
and 7 play hockey and basketball, 5 play cricket and basketball, 4 play hockey
and cricket, 15 do not play any of the three games. Find
(i) How many play hockey, basketball and cricket
 (ii) How many play hockey but not cricket
(iii) How many play hockey and cricket but not basketball

Let's define the sets:

- \( H \): Set of students who play hockey.
- \( B \): Set of students who play basketball.
- \( C \): Set of students who play cricket.

From the problem, we are given the following information:

- Total number of students = 60
- \( |H| = 23 \) (students who play hockey)
- \( |B| = 15 \) (students who play basketball)
- \( |C| = 20 \) (students who play cricket)
- \( |H \cap B| = 7 \) (students who play both hockey and basketball)
- \( |B \cap C| = 5 \) (students who play both basketball and cricket)
- \( |H \cap C| = 4 \) (students who play both hockey and cricket)
- 15 students do not play any of the three games.

Let \( x \) be the number of students who play all three games (hockey, basketball, and cricket). Now let's find:

### (i) How many play hockey, basketball, and cricket:
Using the principle of inclusion and exclusion, we can calculate the total number of students who play at least one of the games:

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

Since 15 students do not play any game, the number of students who play at least one game is:

\[
|H \cup B \cup C| = 60 - 15 = 45
\]

Substitute the known values:

\[
45 = 23 + 15 + 20 - 7 - 5 - 4 + x
\]

Simplify:

\[
45 = 47 - 16 + x
\]

\[
x = 45 - 47 + 16 = 14
\]

Thus, **14 students play all three games**.

### (ii) How many play hockey but not cricket:

To find how many students play hockey but not cricket, we need to subtract those who play both hockey and cricket from the total hockey players:

\[
|H \cap C^c| = |H| - |H \cap C|
\]

But, we must also exclude those who play all three games, as they are part of \( |H \cap C| \). So:

\[
|H \cap C^c| = |H| - (|H \cap C| - |H \cap B \cap C|)
\]

Substitute the known values:

\[
|H \cap C^c| = 23 - (4 - 14) = 23 - 4 + 14 = 9
\]

Thus, **9 students play hockey but not cricket**.

### (iii) How many play hockey and cricket but not basketball:

The number of students who play both hockey and cricket but not basketball is:

\[
|H \cap C \cap B^c| = |H \cap C| - |H \cap B \cap C|
\]

Substitute the known values:

\[
|H \cap C \cap B^c| = 4 - 14 = 3
\]

Thus, **3 students play hockey and cricket but not basketball**.

### Summary of answers:
1. **14 students** play hockey, basketball, and cricket.
2. **9 students** play hockey but not cricket.
3. **3 students** play hockey and cricket but not basketball.

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Do you want more details on any of these steps, or any further clarifications?

Here are 5 related questions:
1. How would the solution change if 10 students do not play any of the games?
2. What is the total number of students who play exactly one game?
3. How many students play basketball but not hockey or cricket?
4. How many students play exactly two of the three games?
5. How many students play neither hockey nor basketball?

**Tip**: The principle of inclusion and exclusion is powerful when solving problems involving multiple sets with overlapping elements.

In a class of 60 students, 23 play hockey, 15 play basketball, 20 play cricket and 7 play hockey and basketball, 5 play cricket and basketball, 4 play hockey and cricket, 15 do not play any of the three games. Find (i) How many play hockey, basketball and cricket (ii) How many play hockey but not cricket (iii) How many play hockey and cricket but not basketball.

Solve a problem involving 60 students, where 23 play hockey, 15 play basketball, and 20 play cricket. Learn how to use the principle of inclusion and exclusion to find how many students play all three games, how many play hockey but not cricket, and more. Suitable for Grades 9-12.

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