Answer :

Here we are given that

A = {4, 5, 6}

B = {5, 6, 7, 8}

C = {6, 7, 8, 9}

Here we have to verify the eq^{n} given below

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)

Here we have to find the cardinality of all the sets and their intersection. So basing on the data given we can find the values for the eq^{n} .

(A Ո B Ո C) will give elements which are common to all the three sets while (A ⋃ B ⋃ C) will give combination of all distinct elements

A = {4, 5, 6}

B = {5, 6, 7, 8}

C = {6, 7, 8, 9}

(A Ո B) = {4, 5, 6} Ո {5, 6, 7, 8} = {5, 6}

(B Ո C) = {5, 6, 7, 8} Ո {6, 7, 8, 9} = {6, 7, 8}

(A Ո C) = {4, 5, 6} Ո {6, 7, 8, 9} = {6}

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

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

= {4, 5, 6, 7, 8, 9}

So from the above expression we can find the cardinality of all the values,

n(A) = 3

n(B) = 4

n(C) = 4

n(AՈB) = 2

n(BՈC) = 3

n(AՈC) = 1

n(A Ո B Ո C) = 1

n(A ⋃ B ⋃ C) = 6

now using the formula,

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)

⇒ 6 = 3 + 4 + 4 – 2 – 3 – 1 + 1

⇒ 6 = 6 hence verified

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