**Formula for A union B union C :**

Here we are going to see the formula for (A U B U C).

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

Let us look into some example problems based on the above formula.

**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)

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

n (A) = 3

n (B) = 4

n (C) = 4

A n B = {4, 5, 6} n {5, 6, 7, 8} = {5, 6}

n (A n B) = 2

B n C = {5, 6, 7, 8} n {6, 7, 8, 9} = {6, 7, 8}

n (B n C) = 3

C n A = {6, 7, 8, 9} n {4, 5, 6} = {6}

n (C n A) = 1

A n B n C = {6}

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

n (A) = 5

n (B) = 3

n (C) = 3

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

n (A n B) = 0

B n C = {x, y, z} n {a, e, x} = {x}

n (B n C) = 1

C n A = {a, e, x} n {a, b, c, d, e} = {a, e}

n (C n A) = 2

A n B n C = { }

n (A n B n C) = 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

After having gone through the stuff given above, we hope that the students would have understood "Formula for a union b union c".

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