A = {b,c,e,g,h}
B = {a,c,d,g,i}
C = {a,d,e,g,h}
B ∩ C = {a,d,g}
A - (B ∩ C) = {b,c,e,g,h} - {a,d,g} = {b,c,e,h}...(1)
A-B = {b,c,e,g,h} - {a,c,d,g,i} = {b,e,h}
A-C = {b,c,e,g,h} - {a,d,e,g,h} = {b,c}
(A - B)⋃(A - C) = {b,c,e,h}...(2)
From (1) and (2) it is verified that
A-(B ⋂ C) = (A - B) U (A - C)

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).

Given, A = {b,e,f,g} and B = {c,e,g,h}
(i) \(A\cup B\) = {b,c,e,f,g,h}     .....(1)
\(B\cup A\) = {b,c,e,f,g,h}        ......(2)
From (1) and (2) we have \(A\cup B\) = \(B\cup A\)
It is verified that union of sets is commutative.
(ii) \(A\cap B\) = {e,g}  .....(3)
\(B\cap A\) = {e,g}   .....(4)
From (3) and (4) we get, \(A\cap B\) = \(B\cap A\)
It is verified that intersection of sets is commutative.

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

n(C) = \(800\times\frac{1}{2}=400\)

n(A∩B) = 70
n(B∩C) = 55
n(C∩A) = 60

(i) Number of people who use exactly two branded soaps
= 50 + 35 + 40 = 125
(ii) Number of people who use atleast one branded soaps
= 190 + 55 + 305 + 50 + 35 + 40 + 20 = 695
(iii) Number of people who do not use any of these brands
= 800 - [695] = 105