Question

Of the members of three athletic teams in a certain school, 21 are on the basketball team, 26 on hockey team and 29 on the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all ?

Answer 1

Let B, H, F denote the sets of members who are in the basket hall team, hockey team and football team respectively.
Given $n\left(B\right)=21,n\left(H\right)=26,n\left(F\right)=29.$ $n(H\cap B)=14,n(H\cap F)=15,$ $n(F\cap B)=12$ and $n(B\cap H\cap F)=8.$
We have to find $n(B\cup H\cup F).$
$n(B\cup H\cup F)=n\left(B\right)+n\left(H\right)n\left(F\right)$ $-n(B\cap H)-n(H\cap F)-n(F\cap B)+n(B\cap H\cap F).$
$n(B\cup H\cup F)=(21+26+29)-(14+15+12)+8=43$