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Q.
In a class of $60$ students, $25$ students play cricket and $20$ students play tennis, and $10$ students play both the games. Then, the number of students who play neither is
Let student play cricket $=C$
Student play tennis $=T$
and total number of students $=S$
$\therefore n(S) =60, n(C)=25,\, n(T)=20$
and $n(C \cap T) =10$
Now, $n(C \cup T) =n(C)+n(T)-n(C \cap T)$
$=25+20-10=35$
$\therefore $ The number of students who play neither game
$=n(C \cap T)'=n(S)-n(C \cup T)$
$=60-35=25$