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 football as well as cricket?

### Concept & Strategy The key insight for solving 3-set Venn diagram problems is distinguishing between "playing exactly two games" and "playing two games." When asked for children who play "football as well as cricket," it includes everyone in the intersection of the Football and Cricket sets, regardless of whether they also play badminton. $$n(F \cap C) = n(\text{Only } F \text{ and } C) + n(\text{All three})$$ ### Step-by-Step Solution * **Given:** * Children playing *only* cricket and football = $4$ * Children playing *all three* games = $2$ * **Calculation:** * The phrase "football as well as cricket" translates to the entire overlapping region between the football and cricket circles. * This region is composed of two distinct parts: those who play *only* those two, and those who play *all three*. * Add the two values together: $4 + 2 = 6$. ### Exam Strategy & Shortcut Instead of drawing a complete Venn diagram, scan the text specifically for the intersection requested. Look for the "only [Game A] and [Game B]" value, then immediately add the "all three" value. ### Common Pitfall The most common mistake is assuming "football as well as cricket" means *only* football and cricket, leading students to select $4$. Always remember that "A and B" inherently includes "A, B, and C" unless the word "only" is explicitly used. ### Final Answer **Therefore, the correct answer is 6.**