Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basket ball 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all?
Let A, B & C be the sets of members in basketball team, hockey team & football team, respectively.
$n(A \cap B)=14$
$n(B \cap C)=15$
$n(A \cap C)=12$
$n(A \cap B \cap C)=8$
$n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(A \cap C)+n(A \cap B \cap C)$
$\Rightarrow n(A \cup B \cup C)=21+26+29-14-15-12+8=43$
Therefore, there are 43 members in all teams.