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

A = {x : x = 2ⁿ, n ∈ W, n < 4}
x = 2⁰ = 1
x = 2¹ = 2
x = 2² = 4
x = 2³ = 8
ஃ A = {1,2,4,8}
B = {x : x = 2n, n∈N and n ≤ 4}
x = 2 \(\times\) 1 = 2
x = 2 \(\times\) 2 = 4
x = 2 \(\times\) 3 = 6
x = 2 \(\times\)4 = 8
ஃ B = {2,4,6,8}
C = {0,1,2,5,6}
Associative property of intersection of sets
A∩(B∩C) = (A∩B)∩C
B∩C = {2,6}
A∩(B∩C) = {1,2,4,6,8}∩{2,6}
= {2} .............(1)
A∩B = {1,2,4,8} ∩ {2,4,6,8} = {2,4,8}
(A∩B) ∩C = {2,4,8} ∩ {0,1,2,5,6}
= {2} ............(2)
From (1) and (2), It is verified that A∩(B∩C) = (A∩B)∩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

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 \(A\cup (B\cup C)\) = \((A\cup B)\cup C\).

From (1) and (2), we get A -(BUC) = (A-B)∩(A-C). Hence it is verified.

\(A\cap B\) = {5,7}
\(B\cap A\)= {5,7}
From (3) and (4) we get, \(A\cap B=B\cap C\)
It is verifield that insersection of sets is commutative.

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)

n(U) = 125
y = 2x
z = x + 10
ஃ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 - 45
4x = 80
x = 20
ஃ y = 2x = 2 \(\times\) 20 = 40
z = x +10 = 20 + 10 = 30
Hence x = 20; y = 40; z = 30

From Venn diagram,
\(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\) is verified

K = {a,b,d,e,f}, L = {b,c,d,g} and M = {a,b,c,d,h}
(i) K\(\cup\)(L\(\cap \)M)
L\(\cap \) M = {b,c,d,g} \(\cap \) {a,b,c,d,h} = {b,c,d}
K\(\cup\)(L\(\cap \)M) = {a,b,d,e,f} \(\cup\) {b,c,d} = {a,b,c,d,e,f}
(ii) K∩(L\(\cup\)M)
L\(\cup\)M = {a,b,c,d,g,h}
K⋂(L \(\cup\) M) = {a,b,d,e,f) \(\cap \) {a,b,c,d,g,h} = {a,b,d}
(iii) (K \(\cup\) L) \(\cap \) (K\(\cup\)M)
K \(\cup\) L = {a,b,c,d,e,f,g}
K \(\cup\) M = {a,b,c,d,e,f,h}
(K \(\cup\) L) \(\cap \) (K \(\cup\) M) = {a,b,c,d,e,f}
(iv) (K \(\cap \) L) \(\cup\) (K \(\cap \) M)
(K \(\cap \) L) = {b,d}
(K \(\cap \) M) = {a,b,d}
(K \(\cap \) L) \(\cup\) (K \(\cap \) M) = {b,d} \(\cup\) {a,b,d} = {a,b,d}

Civen, A = {b, c, d, e} and B = {b, c, e, g} and C = {a, c, e}
Now \(B\cup C\) ={a, b, c, e, g}
\(A\cup \left( B\cup C \right) \) = {a, b, c, d,e, g}      ....(1)
Then,\(A\cup B\) = {b, c, d, e, g}
\(\left( A\cup B \right) \cup C\) = {a, b, c, d, e, g}     ...(2)
From (1) and (2)  it is verified that
\(A\cup \left( B\cup C \right) \) = \(\left( A\cup B \right) \cup C\)

Given,U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {-3, -1, 1, 3, 5, 7}
and B = {1, 2, 3, 4}
Law (i) \(\left( A\cup B \right) ^{ ' }\) =  \({ A }^{ ' }\cap { B }^{ ' }\)
\(\left( A\cup B \right) ^{ ' }\) = {-2, 0, 6, 8, 9} and                                ...(1)
Then,A' ={-2, 0, 2, 4, 6, 8, 9} and
B ' = {-3, -2, -1, 0, 5, 6, 7, 8, 9}
\({ A }^{ ' }\cap { B }^{ ' }\)  = {-2, 0, 6, 8, 9}
From (1) and (2) it is verified that                              ...(2)
\(\left( A\cup B \right) ^{ ' }\) =\({ A }^{ ' }\cap { B }^{ ' }\)
Law (ii) \(\left( A\cap B \right) ^{ ' }\) = \({ A }^{ ' }\cap { B }^{ ' }\)
Now,\(A\cap B\) = {1, 3}
\(\left( A\cap B \right) ^{ ' }\) = {-3, -2, -1, 0, 2, 4, 5, 6, 7, 8, 9}             ...(3)
Then,\({ A }^{ ' }\cup { B }^{ ' }\)  = {-3, -2, -1, 0,2 ,4, 5, 6, 7, 8, 9}      ...(4)
From (3) and (4) it is verified that
\(\left( A\cap B \right) ^{ ' }\) = \({ A }^{ ' }\cup { B }^{ ' }\)

(i) A-B = {3,4,6}
(ii) B-C = {-1,5,7}
(iii) A'UB'
A'UB'
A' = {1,2,0,-3,5,7,8}
B' = {-3,0,1,2,3,4,5,6}
A'UB' = {-3,0,1,2,3,4,5,6,7,8}
(iv) A'⋂B'
A'⋂B' = {-3,0,1,2}
(v) (BUC)'
BUC = {-3,-2,-1,0,3,5,7,8}
(BUC)' = U-(BUC)
= {-3,-2,-1,0,1,2,3,4,5,6,7,8} - {-3,-2,-1,0,3,5,7,8}
(BUC)' = {1,2,4,6}
(vi) A-(BUC) = {-2,-1,3,4,6} - {-3,-2,-1,0,3,5,7,8} = {4,6}
(vii) A-(B⋂C)
B⋂C = {-2,8}
A-(B⋂C) = {-2,-1,3,4,6} - {-2,8} = {-1,3,4,6}