Question

Of the numbers of three athletic team, $26$ on certain school, $21$ are on the basketball team, $426$ 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 on the basket ball team, hockey and football team respectively.
Then we are 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 $n(B\cup H\cup F)$,
To find this, we use the formula
$n(B\cup H\cup F)=n\left(B\right)+n\left(H\right)+n\left(F\right)$
$-n(B\cap H)-n(H\cap F)-n(F\cap B)+n(B\cap H\cap F)$
i.e. $S_1-S_2+S_3$
Hence $n(B\cup H\cup F)=(21+26+29)-(14+15+12)+8-=43$