Question

Out of 60 children in a colony, it was found that 25 like cricket, 26 like football and 26 like badminton. 9 like both cricket and badminton, 11 like both cricket and football and 8 like both football and badminton. 8 children don't like any of the 3 sports. Then the number of children who like all the 3 sports is:

• A. 14
• B. 3
• C. 11
• D. 10

Hint:

To avoid basic errors, always calculate the size of the union set first by subtracting the "none" category from the total universal population.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem can be resolved using set theory and the Principle of Inclusion-Exclusion for three sets.
Key Formula or Approach:
Let $C, F,$ and $B$ represent the sets of children who like cricket, football, and badminton respectively.
The Principle of Inclusion-Exclusion states:
\[ n(C \cup F \cup B) = n(C) + n(F) + n(B) - n(C \cap F) - n(C \cap B) - n(F \cap B) + n(C \cap F \cap B) \]

Step 2: Detailed Explanation:
Let us determine the values from the problem:
Total children, $n(U) = 60$
Children who do not like any sport = 8
Therefore, the number of children who like at least one sport is:
\[ n(C \cup F \cup B) = 60 - 8 = 52 \]
We are given individual and intersection counts:
\[ n(C) = 25 \]
\[ n(F) = 26 \]
\[ n(B) = 26 \]
\[ n(C \cap F) = 11 \]
\[ n(C \cap B) = 9 \]
\[ n(F \cap B) = 8 \]
Let $x = n(C \cap F \cap B)$ be the number of children who like all three sports.
Substitute these values into the Inclusion-Exclusion equation:
\[ 52 = 25 + 26 + 26 - 11 - 9 - 8 + x \]
\[ 52 = 77 - 28 + x \]
\[ 52 = 49 + x \]
\[ x = 52 - 49 = 3 \]
Therefore, exactly 3 children like all three sports.

Step 3: Final Answer
The correct option is (B).