TOPICS

Miscellaneous

Sets

**Example-1 :-** Show that the set of letters needed to spell “ CATARACT ” and the set of letters needed to spell “ TRACT” are equal.

Let X be the set of letters in “CATARACT”. Then X = { C, A, T, R } Let Y be the set of letters in “ TRACT”. Then Y = { T, R, A, C, T } = { T, R, A, C }. Since every element in X is in Y and every element in Y is in X. It follows that X = Y.

**Example-2 :-** List all the subsets of the set { –1, 0, 1 }.

Let A = { –1, 0, 1 }. The subset of A having no element is the empty set φ. The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three elements of A is A itself. So, all the subsets of A are φ, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}.

**Example-3 :-** Show that A ∪ B = A ∩ B implies A = B .

Let a ∈ A. Then a ∈ A ∪ B. Since A ∪ B = A ∩ B , a ∈ A ∩ B. So a ∈ B. Therefore, A ⊂ B. Similarly, if b ∈ B, then b ∈ A ∪ B. Since A ∪ B = A ∩ B, b ∈ A ∩ B. So, b ∈ A. Therefore, B ⊂ A. Thus, A = B

**Example-4 :-** For any sets A and B, show that P ( A ∩ B ) = P ( A ) ∩ P ( B ).

Let X ∈ P ( A ∩ B ). Then X ⊂ A ∩ B. So, X ⊂ A and X ⊂ B. Let Y ∈ P ( A ) ∩ P ( B ). Then Y ∈ P ( A) and Y ∈ P ( B ). So, Y ⊂ A and Y ⊂ B. Therefore, Y ⊂ A ∩ B, which implies Y ∈ P ( A ∩ B ). This gives P ( A ) ∩ P ( B ) ⊂ P ( A ∩ B). Hence P ( A ∩ B ) = P ( A ) ∩ P ( B ).

**Example-5 :-** A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B, what is the least number that must have liked both products?

Let U be the set of consumers questioned, S be the set of consumers who liked the product A and T be the set of consumers who like the product B. Given that n ( U ) = 1000, n ( S ) = 720, n ( T ) = 450. So n ( S ∪ T ) = n ( S ) + n ( T ) – n ( S ∩ T ) = 720 + 450 – n (S ∩ T) = 1170 – n ( S ∩ T ) Therefore, n ( S ∪ T ) is maximum when n ( S ∩ T ) is least. But S ∪ T ⊂ U implies n ( S ∪ T ) ≤ n ( U ) = 1000. So, maximum values of n ( S ∪ T ) is 1000. Thus, the least value of n ( S ∩ T ) is 170. Hence, the least number of consumers who liked both products is 170

**Example-6 :-** Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?

Let U be the set of car owners investigated, M be the set of persons who owned car A and S be the set of persons who owned car B. Given that n ( U ) = 500, n (M ) = 400, n ( S ) = 200 and n ( S ∩ M ) = 50. Then n ( S ∪ M ) = n ( S ) + n ( M ) – n ( S ∩ M ) = 200 + 400 – 50 = 550 But S ∪ M ⊂ U implies n ( S ∪ M ) ≤ n ( U ). This is a contradiction. So, the given data is incorrect.

**Example-7 :-** A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports ?

Let F, B and C denote the set of men who received medals in football, basketball and cricket, respectively. Then n ( F ) = 38, n ( B ) = 15, n ( C ) = 20, n (F ∪ B ∪ C) = 58 and n (F ∩ B ∩ C) = 3. Therefore, n (F ∪ B ∪ C) = n ( F ) + n ( B ) + n ( C ) – n (F ∩ B) – n (F ∩ C) – n (B ∩ C) + n (F ∩ B ∩ C), gives n (F ∩ B) + n (F ∩ C) + n (B ∩ C) = 18 By Venn diagram, a denotes the number of men who got medals in football and basketball only, b denotes the number of men who got medals in football and cricket only, c denotes the number of men who got medals in basket ball and cricket only and d denotes the number of men who got medal in all the three. Thus, d = n ( F ∩ B ∩ C ) = 3 and a + d + b + d + c + d = 18. Therefore a + b + c = 9, which is the number of people who got medals in exactly two of the three sports.

CLASSES