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

We can solve this problem using the principle of inclusion-exclusion. Let's define the sets as follows:

- Let \( H \) be the set of students who play hockey.
- Let \( B \) be the set of students who play basketball.
- Let \( C \) be the set of students who play cricket.
- Let \( n(H) = 23 \), \( n(B) = 15 \), \( n(C) = 20 \).
- Let \( n(H \cap B) = 7 \), \( n(B \cap C) = 5 \), \( n(H \cap C) = 4 \).
- Let \( n(H \cap B \cap C) = x \) (the number of students who play all three sports).

We are also told that 15 students do not play any of the three games. So the total number of students playing at least one game is:

\[
n(H \cup B \cup C) = 60 - 15 = 45
\]

### Step 1: Apply inclusion-exclusion principle
The inclusion-exclusion principle for three sets is:

\[
n(H \cup B \cup C) = n(H) + n(B) + n(C) - n(H \cap B) - n(B \cap C) - n(H \cap C) + n(H \cap B \cap C)
\]

Substitute the given values:

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

Simplify the equation:

\[
45 = 58 - 16 + x
\]

\[
45 = 42 + x
\]

Solve for \( x \):

\[
x = 3
\]

Thus, 3 students play all three sports (hockey, basketball, and cricket).

### Step 2: Answer the questions

(i) **How many play hockey, basketball, and cricket?**
\[
\boxed{3}
\]

(ii) **How many play hockey but not cricket?**

The number of students who play hockey but not cricket is given by:

\[
n(H) - n(H \cap C) = 23 - 4 = 19
\]

So, \( \boxed{19} \) 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:

\[
n(H \cap C) - n(H \cap B \cap C) = 4 - 3 = 1
\]

Thus, \( \boxed{1} \) student plays hockey and cricket but not basketball.

### Summary of answers:
1. 3 students play hockey, basketball, and cricket.
2. 19 students play hockey but not cricket.
3. 1 student plays hockey and cricket but not basketball.

Would you like more details or explanations?

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Here are five related questions to expand on this topic:
1. How many students play exactly two sports?
2. How many students play only basketball?
3. How many students play only one sport?
4. How many students play exactly one or two sports?
5. What percentage of the class plays at least one sport?

**Tip**: In problems involving multiple sets, the inclusion-exclusion principle helps to avoid double-counting when calculating unions of overlapping sets.

In a class of 60 students, 23 play hockey, 15 play basketball, 20 play cricket, 7 play hockey and basketball, 5 play cricket and basketball, 4 play hockey and cricket, and 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 set theory problem involving 60 students playing hockey, basketball, and cricket using the inclusion-exclusion principle. Find how many students play all three sports, hockey but not cricket, and both hockey and cricket but not basketball. Suitable for Grades 10-12.

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