Question

Of the number of three athletic teams in a school, $21$ are in the basketball team, $26$ in hockey team and $29$ in the football team. $14$ play hockey and basketball, $15$ play hockey and football, $12$ play football and basketball and $8$ play all the games. The total number of members is

• A. $42$
• B. $43$
• C. $45$
• D. $46$

Answer 1

Answer: (B)

$43$

Answer 2

Let $B$, $H$ and $F$ be the sets of members in the basketball team, hockey team and football team respectively. $\therefore n(B) = 21$, $n(H) = 26$, $n(F) = 29$, $n(H \cap B) = 14$, $n(H \cap F) = 15$, $n(F \cap B ) = 12$, $n(B \cap H \cap F) = 8$ $\therefore n(B \cup H \cup E) = n(B) + n(H) + n(F) - n(B \cap H)$ $- n(H \cap F) - n(B \cap F) + n(B \cap H \cap F)$ $= 21 + 26 + 29 - 14 - 15 - 12 + 8 = 43$.