FORMULA FOR A UNION B UNION C

n(AuBuC) :

=n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

Explanation :

Let us come to know about the following terms in details.

n(AuB) = Total number of elements related to any of the two events A & B.

n(AuBuC) = Total number of elements related to any of the three events A, B & C.

n(A) = Total number of elements related to A

n(B) = Total number of elements related to B

n(C) = Total number of elements related to C

Example 1 :

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

Solution :

n(AuBuC) ;

= n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(CnA) + n(AnBnC)

= 17 + 17 + 17 - 7 - 6 - 5 + 2

= 35

Example 2 :

A = {4, 5, 6}, B = {5, 6, 7, 8} and C = {6, 7, 8, 9} find the value of n(AuBuC).

Solution :

n(A) = 3

n(B) = 4

n(C) = 4

AnB = {4, 5, 6} n {5, 6, 7, 8}

= {5, 6}  

n(AnB) = 2

BnC = {5, 6, 7, 8} {6, 7, 8, 9} =  {6, 7, 8}

n(BnC) = 3

CnA = {6, 7, 8, 9} n {4, 5, 6} = {6}

n(CnA) = 1

AnBnC = {6}

n(AnBnC) = 1

n(AuBuC) ;

= n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(CnA) + n(AnBnC)

= 3 + 4 + 4 - 2 - 3 - 1 + 1

= 12 - 6

= 6

Example 3 :

A = {a, b, c, d, e}, B = {x, y, z} and C = {a, e, x}  find the value of n(AuBuC).

Solution :

n(A) = 5

n(B) = 3

n(C) = 3

AnB = {a, b, c, d, e} n {x, y, z} = { }

n(AnB) = 0

BnC = {x, y, z} {a, e, x} = {x}

n(BnC) = 1

CnA = {a, e, x} n {a, b, c, d, e} = {a, e}

n(CnA) = 2

AnBnC = { }

n(AnBnC) = 0

n(AuBuC) ;

= n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(CnA) + n(AnBnC)

= 5 + 3 + 3 - 0 - 1 - 2 + 0

= 11 - 3

= 8

Example 4 :

In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. Find the total number of students in the group (Assume that each student in the group plays at least one game).

Solution :

Step 1 :

Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.

Step 2 :

From the given information, we have

n(F) = 65, n(H) = 45, n(C) = 42,

n(FnH) = 20, n(FnC) = 25, n(HnC) = 15

n(FnHnC) = 8

Step 3 :

From the basic stuff, we have

Total number of students in the group is n(FuHuC).

n(FuHuC) is equal to

= n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC)

n(FuHuC) = 65 + 45 + 42 -20 - 25 - 15 + 8

n(FuHuC) = 100

Hence, the total number of students in the group is 100.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 72)

    Nov 23, 24 09:36 PM

    digitalsatmath57.png
    Digital SAT Math Problems and Solutions (Part - 72)

    Read More

  2. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Nov 23, 24 10:01 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 76)

    Nov 23, 24 09:45 AM

    digitalsatmath63.png
    Digital SAT Math Problems and Solutions (Part - 76)

    Read More