Among a group of students, 50 played Cricket, 50 played hockey and 40 played Volley ball, 5 played both cricket and hockey, 10 played both hockey and Volleyball, 5 played Cricket and Volleyball and 10 played all three. If every student played at least one game, find the number of students and how many Played only cricket, only hockey and only Volley ball.

We can solve this problem using the principle of inclusion and exclusion. Let’s define the sets:

- Let \( C \) be the set of students who play Cricket.
- Let \( H \) be the set of students who play Hockey.
- Let \( V \) be the set of students who play Volleyball.

Given dataWe can solve this problem using the principle of inclusion and exclusion. Let’s define the sets:

- Let \( C \) be the set of students who play Cricket.
- Let \( H \) be the set of students who play Hockey.
- Let \( V \) be the set of students who play Volleyball.

Given 
- \( |C| = 50 \) (students who played Cricket),
- \( |H| = 50 \) (students who played Hockey),
- \( |V| = 40 \) (students who played Volleyball),
- \( |C \cap H| = 5 \) (students who played both Cricket and Hockey),
- \( |H \cap V| = 10 \) (students who played both Hockey and Volleyball),
- \( |C \cap V| = 5 \) (students who played both Cricket and Volleyball),
- \( |C \cap H \cap V| = 10 \) (students who played all three games).

### Step 1: Total number of students
Using the principle of inclusion and exclusion, the total number of students \( N \) is:

\[
N = |C| + |H| + |V| - |C \cap H| - |H \cap V| - |C \cap V| + |C \cap H \cap V|
\]

Substituting the values:

\[
N = 50 + 50 + 40 - 5 - 10 - 5 + 10 = 130 - 20 + 10 = 120
\]

So, the total number of students is \( 120 \).

### Step 2: Students who played only one game

#### Only Cricket:
Students who played only Cricket are those in \( C \) but not in \( H \) or \( V \). This is calculated as:

\[
\text{Only Cricket} = |C| - |C \cap H| - |C \cap V| + |C \cap H \cap V|
\]

Substituting the values:

\[
\text{Only Cricket} = 50 - 5 - 5 + 10 = 50 - 10 + 10 = 50
\]

So, \( 50 \) students played only Cricket.

#### Only Hockey:
Students who played only Hockey are those in \( H \) but not in \( C \) or \( V \). This is calculated as:

\[
\text{Only Hockey} = |H| - |C \cap H| - |H \cap V| + |C \cap H \cap V|
\]

Substituting the values:

\[
\text{Only Hockey} = 50 - 5 - 10 + 10 = 50 - 15 + 10 = 45
\]

So, \( 45 \) students played only Hockey.

#### Only Volleyball:
Students who played only Volleyball are those in \( V \) but not in \( C \) or \( H \). This is calculated as:

\[
\text{Only Volleyball} = |V| - |C \cap V| - |H \cap V| + |C \cap H \cap V|
\]

Substituting the values:

\[
\text{Only Volleyball} = 40 - 5 - 10 + 10 = 40 - 15 + 10 = 35
\]

So, \( 35 \) students played only Volleyball.

### Final Answer:
- Total number of students: \( 120 \)
- Number of students who played only Cricket: \( 50 \)
- Number of students who played only Hockey: \( 45 \)
- Number of students who played only Volleyball: \( 35 \)

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Would you like further details or have any questions? Here are some related questions you might explore:

1. How many students played exactly two games?
2. How many students played exactly one game?
3. How many students played all three games?
4. If 5 more students played only Hockey, how would the total change?
5. What is the minimum number of students who must have played two games?

**Tip:** The inclusion-exclusion principle is very useful in counting problems involving multiple overlapping sets.

Solve a problem using the Inclusion-Exclusion Principle involving 50 students playing Cricket, 50 playing Hockey, and 40 playing Volleyball. Learn how to calculate the total number of students and those who played only one sport. Suitable for Grades 10-12.

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