## Set Theory Practice Solution5

In this page set theory practice solution5 we are going to see solution of practice questions from the worksheet set theory practice questions2.

Question 4:

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:

To verify the condition n (AUBUC)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩ B ∩C) we have to find number of terms in A,B,C number of terms in A U B ,B U C and C U A and also number of terms in (A∩ B ∩C)

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)

= 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) = 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)

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

= 11 - 3

n (A U B U C) = 8

Question 5:

In a college,60 students enrolled in chemistry,40 in physics,30 in biology,15 in chemistry and physics,10 in physics and biology,5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.

Solution:

Let A,B and C are the sets enrolled in the subjects Chemistry,Physics  and Biology respectively.

Number of students enrolled in Chemistry n (A) = 60

Number of students enrolled in Physics n (B) = 40

Number of students enrolled in Biology n (C) = 30

Number of students enrolled in Chemistry and Physics n (A ∩ B) = 15

Number of students enrolled in Physics and Biology n (B ∩ C) = 10

Number of students enrolled in Biology and Chemistry n (C ∩ A) = 5

No one enrolled in all the three,So n (A ∩ B ∩ C) = 0 Number of students enrolled

in at least one of the subjects = 35 + 15 + 20 + 10 + 5 + 15

= 100 set theory practice solution5 set theory practice solution5

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