Question

If n(A UBUC) + n(AU BUC)' = 50, n(A) = 10, n(B) = 15, n(C) = 20, n(An B) = 5, n(B in C) = 6, n(A) = 10 and n(
ABC) = 5, then
n(A UBUC)' is equal to
O 21​

Answer 1

Step-by-step explanation:

QUESTION :

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}

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

CARDINAL NUMBER OF A FINITE SET :

The Cardinal number of a finite set A is the number of distinct elements in the set A. It is denoted by n(A).

•It is not possible to define the Cardinal number of an infinite set.

•The Cardinal number of the empty set is zero.

•The Cardinal number of a singleton set is 1.

Intersection of two sets :

The intersection of the sets a and b is the set of all the elements which belong to both A and B. It is denoted by A ∩ B (“ A intersection B”).

•If A and B do not have any element in common then A ∩ B= a null set = Ø.

SOLUTION :

First ,we have to find number of terms in A U B ,B U C and C U A and also number of terms in (A∩ B ∩C)

GIVEN :

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) = {5,6}

n (A ∩ B) = 2

(B ∩ C) = {6,7,8}

n (B ∩ C) = 3

(A ∩ C) ={6}

n (A ∩ C) = 1

(A ∩ B ∩ C) = {6}

n (A ∩ B ∩ C) =1

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

n (A U B U C) = 3 + 4 + 4 - 2 - 3 - 1 + 1

n (A U B U C) = 11 - 6 + 1

n (A U B U C) = 12 - 6

n (A U B U C) = 6

(ii)

GIVEN :

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) = 0

B ∩ C = {x}

n (B ∩ C) = 1

C ∩ A = {a,e}

n (C ∩ A) = 2

n (A ∩ B ∩ C) = 0

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

n (A U B U C) = 5 + 3 + 3 - 0 - 1 - 2 + 0

n (A U B U C) = 11 - 3

n (A U B U C) = 8

Hence,n (A U B U C) = 8

HOPE THIS WILL HELP YOU...