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

Let B, H, F denote the sets of members who are on the basketball team, hockey team and football team respectively.

Then we are given $\mathrm{n}\left(\mathrm{B}\right)=21, \mathrm{n}\left(\mathrm{H}\right)=26,\mathrm{n}\left(\mathrm{F}\right)=29$

$\mathrm{n}(\mathrm{H}\cap \mathrm{B})=14, \mathrm{n}(\mathrm{H}\cap \mathrm{F})=15, \mathrm{n}(\mathrm{F}\cap \mathrm{B})=12$

and $\mathrm{n}(\mathrm{B}\cap \mathrm{H}\cap \mathrm{F})=8.$

We have to find $\mathrm{n}(\mathrm{B}\cup \mathrm{H}\cup \mathrm{F}).$

To find this, we use the formula

$\mathrm{n}(\mathrm{B}\cup \mathrm{H}\cup \mathrm{F})=\mathrm{n}\left(\mathrm{B}\right)+\mathrm{n}\left(\mathrm{H}\right)+\mathrm{n}\left(\mathrm{F}\right)$

$-\mathrm{n}(\mathrm{B}\cap \mathrm{H})-\mathrm{n}(\mathrm{H}\cap \mathrm{F})-\mathrm{n}(\mathrm{F}\cap \mathrm{B})+\mathrm{n}(\mathrm{B}\cap \mathrm{H}\cap \mathrm{F})$

Hence, $\mathrm{n}(\mathrm{B}\cup \mathrm{H}\cup \mathrm{F})=(21+26+29)-(14+15+12)+8=43$

Thus these are 43 members in all.