Question

Out of the members of three athletic team in a school, $21$ are in the basket-ball team, $26$ in the 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?

• A. $41$
• B. $42$
• C. $43$
• D. $44$

Answer 1

The correct option is C $43$

Let,

$H$ denote the set of hockey players

.

$B$ denote the set of basket players.

And $F$ denote the set of football players.

Given $n\left(H\right)=26,n\left(B\right)=21,n\left(F\right)=29,n(H\cap B)=14,n(F\cap B)=15,n(H\cap F)=12,n(H\cap b\cap F)=8$

Hence,

$n(H\cup B\cup F)=nH+nB+nF-n(H\cap B)-n(F\cap B)-n(H\cap F)+n(H\cap B\cap F)$

$=26+21+29-(14+15+12)+8$

$=76-\left(41\right)+8$

$=35+8$

$=43$