FORMULA FOR A UNION B UNION C

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

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.

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

About Us  |  Contact Us  |  Privacy Policy

©All rights reserved. onlinemath4all.com

onlinemath4all_official_badge1.png

Recent Articles

  1. 10 Tricky SAT Math Questions with Answers

    Feb 25, 26 08:07 AM

    10 Tricky SAT Math Questions with Answers

    Read More

  2. Solving Exponential Equations

    Feb 23, 26 10:06 AM

    exptolog.png
    Solving Exponential Equations - Concept - Examples

    Read More

  3. SAT Math Problems on Exponents and Radicals

    Feb 21, 26 08:28 PM

    SAT Math Problems on Exponents and Radicals

    Read More