Out of 450 students of a school, 325 play football, 175 play cricket and 50 neither play football nor cricket. How many students play both football and cricket?

### Concept & Formula To find the intersection of two sets when a portion of the total population belongs to neither set, first subtract the "neither" group from the total population to find the union. Then, apply the Inclusion-Exclusion formula. $$ \text{Total} = n(A \cup B) + \text{Neither} $$ $$ n(A \cup B) = n(A) + n(B) - n(A \cap B) $$ ### Step-by-Step Solution * **Given:** * Total students = $450$ * Play football, $n(F) = 325$ * Play cricket, $n(C) = 175$ * Play neither = $50$ * **Calculation:** First, determine how many students play at least one of the games ($n(F \cup C)$). $n(F \cup C) = \text{Total} - \text{Neither}$ $n(F \cup C) = 450 - 50 = 400$ Now, use the set theory formula to find the number of students who play both ($n(F \cap C)$). $n(F \cup C) = n(F) + n(C) - n(F \cap C)$ $400 = 325 + 175 - n(F \cap C)$ $400 = 500 - n(F \cap C)$ $n(F \cap C) = 500 - 400$ $n(F \cap C) = 100$ ### Exam Strategy & Shortcut Sum the individuals for the distinct groups and the "neither" group: $325 \text{ (Football)} + 175 \text{ (Cricket)} + 50 \text{ (Neither)} = 550$. Compare this sum to the actual total number of students ($450$). The excess amount is exactly the number of students who were counted twice (i.e., those who play both). Excess = $550 - 450 = 100$. ### Common Pitfall Forgetting to account for the $50$ students who play neither sport. If you assume the union is the total population ($450$), you will erroneously calculate $325 + 175 - 450 = 50$, leading directly to trap option (a). ### Final Answer Therefore, the correct answer is **100**.