Directions: These questions are based on the following information: Children in a class play only one or two or all of the three games - badminton, football and cricket. 5 children play only cricket, 8 children play only football and 7 children play only badminton. 3 children play only two games - badminton and football, 4 children play only two games - cricket and football, and another 4 children play only two games - badminton and cricket. 2 children play all the three games. How many children play badminton as well as cricket?

### Concept & Strategy When calculating the intersection of two specific sets, you must combine the region where *only* those two sets intersect with the central region where *all three* sets intersect. $$n(B \cap C) = n(\text{Only } B \text{ and } C) + n(\text{All three})$$ ### Step-by-Step Solution * **Given:** * Children playing *only* badminton and cricket = $4$ * Children playing *all three* games = $2$ * **Calculation:** * The question asks for "badminton as well as cricket" without the restrictive word "only". * This means we must include everyone who plays these two games, even if they also happen to play football. * Add the dual-sport group to the tri-sport group: $4 + 2 = 6$. ### Exam Strategy & Shortcut Identify the two games in question (Badminton and Cricket). Find the "only two games" value for this pair ($4$), and automatically add the universal "all three" value ($2$) to it. ### Common Pitfall Forgetting to add the center intersection ($2$) is the most frequent mistake. Students see "only two games - badminton and cricket = 4" and immediately choose $4$. The absence of the word "only" in the question itself changes everything. ### Final Answer **Therefore, the correct answer is 6.**