PROBLEMS ON INTERSECTION OF THREE SETS

For solving such a problems we have to consider the following rules :

If A, B and C are three finite sets then :

n(AUBUC) 

=  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

Question 1 :

For any three sets A,B and C if n(A) = 17, n(B) = 17, n(C) =  17, n(A∩B)  =  7, n(B∩C)  =  6, (A∩C)  =  5 and n(A∩B∩C) = 2, find n (AUBUC).

Solution :

n(A)  =  17

n(B)  =  17

n(C)  =  17

n(A∩B)  =  7

n(B∩C)  =  6

(A∩C)  =  5 

n(AUBUC)

= n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n (A∩B∩C)

  =  17 + 17 + 17 - 7 - 6 - 5 + 2

  =  53 - 18 + 2

  =  55 - 20

n (AUBUC)  =  35

Question 2 :

verify

n(AUBUC)

  =  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩B ∩C)

(i) A  =  {4, 5, 6}, B  =  {5, 6, 7, 8} and C  =  {6, 7, 8, 9}

Solution :

A  =  {4, 5, 6}

B  =  {5, 6, 7, 8}

C  =  {6, 7, 8, 9}

n(A)  =  3, n(B)  =  4, n(C)  =  4

A∩B∩C  =  {6}, n(A∩B∩C)  =  1

n(AUBUC)

  =  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

  =  3 + 4 + 4 - 2 - 3 - 1 + 1

  =  11 - 6 + 1

  =  12 - 6

  =  6

(ii) A  =  {a, b, c, d, e} B  =  {x, y, z} and C = {a, e, x}

Solution :

A  =  {a, b, c, d, e} B  =  {x, y, z} and C  =  {a, e, x}

n(A)  =  5 , n(B)  =  3, n(C)  =  3

n (A∩B∩C)  =  0

n(AUBUC)

  =  n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)

  =  5 + 3 + 3 - 0 - 1 - 2 + 0

  =  11 - 3

 n (AUBUC)  =  8 

Related pages

• Back to worksheet
  
• Definition
• Representation of Set 
• Types of set
• Disjoint sets 
• Power Set
  
• Operations on Sets
• Laws on set operations
• More Laws
• Venn diagrams
• Set word problems
• Relations and functions

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