Question

Of the members of 3 athletic teams in a certain school, 21 are in the basketball team, 26 in the hockey team and 29 in 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

Hint: To find total members in all the teams, we have to find the union of members in the basketball, hockey and football teams. Let us assume that the number of the members in the basketball, hockey and football teams is n (B), n (H) and n (F) respectively. Now, use the formula: n (B∪H∪F) = n (B) + n (H) + n (F) + n (B∩H) + n (H∩F) + n (F∩B) – n (B∩H∩F).

Complete step-by-step answer:
Let us assume that:

Number of members in the basketball team is n (B).

Number of members in the hockey team is n (H).

Number of members in the football team is n (F).

In the question above, we have asked to find the total members in all means we have to find the union of the members in all the three teams which is found by using the below formula:

n (B∪H∪F) = n (B) + n (H) + n (F) + n (B∩H) + n (H∩F) + n (F∩B) – n (B∩H∩F)………… Eq. (1)
In the above formula,

n (B∩H) represents the number of members that play both basketball and hockey.

n (H∩F) represents the number of members that play both hockey and football.

n (F∩B) represents the number of members that play both football and basketball.

n (B∩H∩F) represents the number of members that play both football, hockey and basketball.

It is given that,

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

n (B∩H) = 14, n (H∩F) = 15, n (F∩B) = 12, n (B∩H∩F) = 8

Now, substituting the above values in eq. (1) we get,

n (B∪H∪F) = n (B) + n (H) + n (F) + n (B∩H) + n (H∩F) + n (F∩B) – n (B∩H∩F)

$\Rightarrow $ n (B∪H∪F) = 21 + 26 + 29 + 14 + 15 + 12 – 8

$\Rightarrow $ n (B∪H∪F) = 109

Hence, the total members which are there in all are 109.

Note: Some things that need to be careful about while writing the formula for n (B∪H∪F) is that:

Don’t forget to subtract n (B∩H∩F) in this formula n (B∪H∪F).

Make sure to write all these terms in this expression “n (B∩H) + n (H∩F) + n (F∩B)”.