\(n\left( A\cup B\cup C \right) \)= n(A) + n(B) + n(C) - \(n\left( A\cap B \right) -\left( B\cap C \right) -n\left( A\cap C \right) +\left( A\cap B\cap B \right) \)

For any three sets A, B and C
(i) \(A\cap \left( B\cap C \right) =\left( A\cap B \right) \cap C\)
(ii) \(A\cup \left( B\cup C \right) =\left( A\cup B \right) \cup C\)

Given A = {0, 2, 4,6, 8} , B = {2, 3, 5, 7} and C = {5, 6, 7, 8}
First, we find \(B\cap C\) = {5, 7} ,\(A\cup (B\cap C)\) = {0, 2, 4, 5, 6, 7, 8} .......... (1)
Next, \(A\cup B\) = {0,2,3,4,5,6,7, 8},  \(A\cup C\) = {0, 2, 4, 5, 6, 7, 8}
Then, \((A\cup B)\cap (A\cup C)\) = {0, 2, 4, 5, 6, 7, 8} .............(2)
From (1) and (2), it is verified that \((A\cup \left( B\cap C \right) =(A\cup B)\cap (A\cup C)\).

Associative Property of union of sets
AU(BUC) = (AUB)UC
BUC = {m, n, q, s, t} U {m, n, p, q, s} = {m, n, p, q, s, t}
AU(BUC) = {p, q, r, s} U {m, n, p, q, s, t} = {m, n, p, q, r, s, t} ...(1)
(AUB) = {p, q, r, s} U {m, n, q, s, t} = {p, q, r, s, m, n, t}
(AUB)UC = {p, q, r, s, m, n, t} U {m, n, p, q, s} = {p, q, r, s, m, n, t}...(2)
From (1) & (2)
It is verified that AU(BUC) =( AUB)UC

(i) (P∪Q)∪R
(P∪Q) = {1, 2, 5, 7, 9} U {2, 3, 5, 9, 11} = {1,2,3,5,7,9,11}
(P∪Q)∪R = {1,2,3,5,7,9,11} U {3,4,5,7,9} = {1,2,3,4,5,7,9,11}
(ii) (P⋂Q)⋂S
(P⋂Q) = {1,2,5,7,9} ⋂ {2,3,5,9,11} = {2,5,9}
(P⋂Q)⋂S = {2,5,9} ⋂ {2,3,4,5,8} = {2,5}
(iii) (Q⋂S)⋂R
(Q⋂S) = {2,3,5,9,11} ⋂ {2,3,4,5,8} = {2,3,5}
(Q⋂S)⋂R = {2,3,5} ⋂ {3,4,5,7,9} = {3,5}

Given, A = {2, 3, 4, 5},B = {2, 3, 5,7} and C = {1, 3, 5}
Now, \(B\cup C\) = {1,2, 3,5,7}
\(A\cup (B\cup C)\) ={1, 2, 3, 4, 5, 7}        .....(1)
Then, \(A\cup B\) = {2, 3, 4,5,7}
\((A\cup B)\cup C\) = {1,2, 3, 4, 5, 7}    ...(2)
From (1) and (2), it is verified that \(A\cup (B\cup C)\) = \((A\cup B)\cup C\).

Given, A = {2,5,6,7} and B = {3,5,7,8}
\(A\cup B\) = {2,3,5,6,7,8}  ...........(1)
\(B\cup A\) = {2,3,5,6,7,8} ..............(2)

A = {-2,0,1,3,5}, B = {-1,0,2,5,6}
C = {-1,2,5,6,7}
B∪C = {-1,0,2,5,6,7}
A-(B∪C) = {-2,1,3}.............(1)
(A-B) = {-2,1,3}
(A-C) = {-2,0,1,3}
(A-B)∩(A-C) = {-2,1,3}............(2)
From (1) and (2), it is verified that
A-(B∪C) = (A-B)⋂(A-C)

The roster form of sets P, Q and R are
P = {1,2, 3, 4,5,6,7,8,9}, Q = {1, 3,5,7,9} and R = {2, 3, 5,7,11,13}
First, we find (Q ⋂ R) = {3, 5, 7}
Then, P −(Q ⋂ R) = {1, 2, 4 ,6, 8, 9}........ (1)
Next, P–Q = {2, 4, 6, 8} and P–R = {1, 4, 6, 8, 9}
and so, (P −Q) U (P −R) = {1,2, 4, 6, 8, 9}............ (2)
Hence from (1) and (2), it is verified that P −(Q ∩ R) = (P −Q) U (P −R).
Finding the elements of set Q
Given, x = 2n + 1
n = 0 ⟶ x = 2(0) + 1 = 0 + 1 = 1
n = 1 ⟶ x = 2(1) + 1 = 2 + 1 = 3
n = 2 ⟶ x = 2(2) + 1 = 4 + 1 = 5
n = 3 ⟶ x = 2(3) + 1 = 6 + 1 = 7
n = 4 ⟶ x = 2(4) + 1 = 8 + 1 = 9
Therefore, x takes values such as 1, 3, 5, 7 and 9.

(2)=(5)
ஃ (A∩B)'=A'UB'

ஃ A-(B∩C)=(A-B)U(A-C)
Hence it is proved.

A = {11,13,14,15,16,18}
B = {11,12,15,16,17,19}, C = {13,15,16,17,18,20}
(BUC) = {11,12,13,15,16,17,18,19,20}
A⋂(BUC) = {11,13,14,15,16,18}⋂{11,12,13,15,16,17,18,19,20}
= {11,13,15,16,18}...(1)
A⋂B = {11,15,16}; A⋂C = {13,15,16,18}
(A⋂B)U(A⋂C) = {11,13,15,16,18}...(2)
From (1) and (2) it is verified that A⋂(BUC) = (A⋂B)U(A⋂C).

From (1) and (2), \(A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\) is verified.

a = 600 - (120 -x + x + 80 -x)
= 600 - (200 -x)
= 600 - 200 + x = 400 + x
b = 350 - (120 -x + x + 100 -x)
= 350 - (220 - x)
= 350 - 230 + x = 130 + x
c = 280 - (80 - x + x + 100 - x)
280 - (180 -x)
= 280 - 180 + x = 100 + x
Each farmer grew at least one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 - x + 100 - x + 80 - x + x = 1000
400 + x + 130 + x + 100 + x + 120 - x + 100 - x + 80 - x + x = 1000
ஃ 930 + x = 1000
x = 1000 - 930 = 70
ஃ 70 farmers grew all the three crops.

A = {x : x = 6n, n ∈ W, n < 6}
x = 6n
n = {0,1,2,3,4,5}
x = 6 \(\times\) 0 = 0
x = 6 \(\times\) 1 = 6
x = 6 \(\times\) 2 =12
x = 6 \(\times\) 3 =18
x = 6 \(\times\) 4 = 24
x = 6 \(\times\) 5 = 30
ஃ A = {0,6,12,18,24,30}
B = {x : x = 2n, n∈N, 2n = {3,4,5,6,7,8,9}
x = 2n
x = 2 \(\times \) 3 = 6;
2\(\times\)4 = 8;
2 \(\times\)​​​​5 = 10 
2 \(\times\)6 = 12 
2\(\times\) 7 = 14 
2 \(\times\)8 = 16 
2 \(\times\)9 = 18
ஃ B = {6,8,10,12,14,16,18}
C = {x : x = 3n, n∈N, 4 ≤ n < 10
N = {4,5,6,7,8,9}
x = 3\(\times\)4 = 12 
x = 3 \(\times\) 5 = 15 
x = 3 \(\times\) 6 = 18
x = 3 \(\times\) 7 = 21 
x = 3 \(\times\) 8 = 24 
x = 3 \(\times\) 9 = 27
x = 2 \(\times\) 9 = 18
ஃ C = {12,15,18,21,24,27}
A-(B⋂C) = (A-B)U(A-C)
L.H.S = R.H.S
B⋂C = {12,18}
A-(B⋂C) = {0,6,12,18,24,30} - {12,18}
= {0,6,24,30} .........(1)
(A-B) = {0,24,30}
(A-C) = {0,6,30}
(A-B)U(A-C) = {0,6,24,30} .........(2)
From (1) and (2), it is verified that
A-(B⋂C) = (A-B)U(A-C).