Application of SETs | Introduction to SETs, Part V

Continuation from previous posts

1. Introduction to SETs, Part1 
2. Introduction to SETs, Part II
3. Introduction to SETs, Part III
4. Laws of sets

Now we learn some application of sets that helpful in solving many practical problems. We have already familiar with Venn diagrams that are helpful in establishing many important relations between different sets.

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
2. n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A)
3. n(A - B) = n(A) - n(A ∩ B)
4. n(B - A) = n(B) - n(A ∩ B)
5. n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

Based on the above formula let us consider an example.

Example_1: From a Spanish football club’s 30 players, 24 can speak Spanish and 9 can speak English. Then how many can speak
(i) both Spanish and English
(ii) Spanish only
(iii) English only

Solution_1: Let A and B denotes the set of players who can speak Spanish and English, respectively. Then in usual notations, we are given

n(A ∪ B) = 30, n(A) = 24, n(B) = 9

(i) We know that
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
=> 24 + 9 – n(A ∩ B)
=> 33 – 30
=> 3

\ Number of players who can speak both Spanish and English = 3

(ii) We know that
n(A - B) = n(A) - n(A ∩ B)
=> 24 – 3
=> 21

\ Number of players who can speak Spanish only = 21

(iii) We know that
n(B - A) = n(B) - n(A ∩ B)
=> 9 – 3
=> 6

\ Number of players who can speak English only = 6

Example_2: Out of 50 employees in a company, 24 play cricket, 15 play hockey, 18 play football, 6 play cricket and hockey, 8 play cricket and football, 5 play hockey and football and 10 employees do not play any of the three games. Then how may play
(i) All the three games,
(ii) Hockey (H) but not football (F),
(iii) Cricket (C) and football but not hockey

Solution_2: Let C, H, F denotes the set of employees who play cricket, hockey and football respectively. Then in usual notations, we are given

n(C) = 24, n(H) = 15, n(F) = 18, n(C ∩ H) = 6, n(C ∩ F) = 8, n(H ∩ F) = 5, n(C’ ∩ H’ ∩ F’) = 10

(i) First we have to find out the number of employees who play at least one of the three games which is given as

n(C ∪ H ∪ F) = 50 - n(C’ ∩ H’ ∩ F’) = 50 – 10 = 40

Now from application of sets we know that,
n(C ∪ H ∪ F) = n(C) + n(H) + n(F) - n(C ∩ H) - n(C ∩ F) - n(H ∩ F) + n(C ∩ H ∩ F)
=> n(C ∪ H ∪ F) = 24 + 15 + 18 – 6 – 8 – 5 + n(C ∩ H ∩ F)
=> 40 = 38 + n(C ∩ H ∩ F)
=> n(C ∩ H ∩ F) = 40 – 38 = 2

\ Number of employees who play all the three games = 2

(ii) As from application of sets we know that,
n(H - F) = n(H) – n(H ∩ F)
=> 15 – 5
=> 10

\ Number of employees who play hockey but not football = 10

(iii) To find out those employees who play cricket and football but not hockey are,
n((C ∩ F) - H)
= n(C ∩ F) – n((C ∩ F) ∩ H)
= 8 – 2
= 6

\ Number of employees who play cricket and football but not hockey = 6

In my SETs theory articles series I have covered definition of set with examples, methods of writing sets, empty set, singleton set, finite set, infinite set, equivalent set, equal set, subsets, proper subset, super set, universal set, power set, Venn diagrams, various operations on sets, idempotent, identity, commutative, associative, distributive and De-Morgan's laws.