Question

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 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 ?

Answer 1

Let,

n(P) denote total nuber of members,\

n(B) denote number of memebers in the basket ball team

n(H) denote number of members in the hockey team and

n(F) denote number of members in the football team.

Then,

n(B) = 21, n(H) = 26, and n(F) = 29

Also, $n(H\cap B)=14,n(H\cap F)=15,n(F\cap B)=12,n(H\cap B\cap F)=8$

Now,

P = $B\cup H\cup F$

n(P) = $n(B\cup H\cup F)$

= $n\left(B\right)+n\left(H\right)+n\left(F\right)-n(B\cap H)-n$

$(H\cap F)-n(B\cap F)+n(B\cap H\cap F)$

$\Rightarrow n\left(P\right)=21+26+29-14-15-12+8$

= 76-41+8

= 43

Hence, there are 43 members in all.