In a school, all the students play at least one of the games - chess, carrom and table tennis. 60 play chess, 50 play table tennis, 48 play carrom. 12 play chess and carrom. 15 play carrom and table tennis, 20 play table tennis and chess.
What can be the minimum number of students in the school?

Options

A. 123
B. 111
C. 95
D. 63

Answer

To find the minimum number of students in the school who play at least one of the games (chess, carrom, and table tennis), we can use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Define the Sets:**
- Let \( A \) be the set of students who play chess.
- Let \( B \) be the set of students who play carrom.
- Let \( C \) be the set of students who play table tennis.

2. **Given Information:**
- \( |A| = 60 \) (students play chess)
- \( |B| = 48 \) (students play carrom)
- \( |C| = 50 \) (students play table tennis)
- \( |A \cap B| = 12 \) (students play both chess and carrom)
- \( |B \cap C| = 15 \) (students play both carrom and table tennis)
- \( |C \cap A| = 20 \) (students play both table tennis and chess)

3. **Using Inclusion-Exclusion Principle:**
The formula for the total number of students who play at least one game is:
\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C| \]
where \( |A \cap B \cap C| \) is the number of students who play all three games.

4. **Substituting the Values:**
\[ |A \cup B \cup C| = 60 + 48 + 50 - 12 - 15 - 20 + |A \cap B \cap C| \]
Simplifying this gives:
\[ |A \cup B \cup C| = 60 + 48 + 50 - 12 - 15 - 20 + |A \cap B \cap C| \]
\[ = 158 - 47 + |A \cap B \cap C| \]
\[ = 111 + |A \cap B \cap C| \]

5. **Finding Minimum Students:**
To find the minimum number of students, we can assume that no student plays all three games, which means \( |A \cap B \cap C| = 0 \).
Therefore:
\[ |A \cup B \cup C| = 111 + 0 = 111 \]

### Conclusion:
The minimum number of students in the school is **111**.