The following questions are based on the information given below:
Out of a group of 60 students, 25 play Cricket, 30 play Football, 24 play Volleyball, 10 play Cricket and Football, 9 play Cricket and Volleyball, 12 play Volleyball and Football and 5 play all the three.
How many students do not play any one of the games?
A. 5
B. 2
C. 7
D. 1

Options

A. A
B. C
C. B
D. D

Answer

To find out how many students do not play any of the games, we can use the principle of inclusion-exclusion. Let's denote:

- \( n(C) \): Number of students who play Cricket = 25
- \( n(F) \): Number of students who play Football = 30
- \( n(V) \): Number of students who play Volleyball = 24
- \( n(C \cap F) \): Number of students who play both Cricket and Football = 10
- \( n(C \cap V) \): Number of students who play both Cricket and Volleyball = 9
- \( n(F \cap V) \): Number of students who play both Football and Volleyball = 12
- \( n(C \cap F \cap V) \): Number of students who play all three games = 5

Now, we can use the formula for the union of three sets:

\[ n(C \cup F \cup V) = n(C) + n(F) + n(V) - n(C \cap F) - n(C \cap V) - n(F \cap V) + n(C \cap F \cap V) \]

Substituting the values we have:

\[ n(C \cup F \cup V) = 25 + 30 + 24 - 10 - 9 - 12 + 5 \]

Calculating step-by-step:

1. Add the number of students playing each game:
\[ 25 + 30 + 24 = 79 \]

2. Subtract the number of students playing two games:
\[ 79 - 10 - 9 - 12 = 48 \]

3. Add back the number of students playing all three games:
\[ 48 + 5 = 53 \]

So, \( n(C \cup F \cup V) = 53 \).

Now, to find the number of students who do not play any of the games, we subtract the number of students who play at least one game from the total number of students:

\[ \text{Number of students not playing any game} = \text{Total students} - n(C \cup F \cup V) \]
\[ = 60 - 53 = 7 \]

Thus, the number of students who do not play any of the games is **7**.