3.3 Sets Part 2:
Introduction: Is it not interesting to solve a problem similar to:
“A class has 60 students. Every one should
choose to be in Kabadi team or hockey team or in both
the teams. If 45 students chose to be in Kabadi team
and 30 students chose to be in Hockey team, how many are in both the teams”
In this chapter we solve similar
problems.
3.3.1 Properties of sets:
We know that 2+3 =3+2 and 2*3 =3*2.
Thus addition and multiplication are
commutative.
Similarly (2+3)+4=
2+(3+4) and (2*3)*4= 2*(3*4).
Thus addition and multiplication are
associative.
Let us study these properties for sets
3.3.1 Example 1 : Let us
consider the sets A = {p,q,r,} ,B = {q,r,s,} and C={r,s,t}
Prove that
1.
B_{}C =C_{}B
2.
B_{}C = C_{}B
3.
A_{}(B_{}C) = (A_{}B)_{}C
4.
A_{}(B_{}C) = (A_{}B) _{}C
5.
A _{}(B_{}C) = (A_{}B) _{} (A_{}C)
6.
A_{} (B_{}C) = (A_{}B)_{}(A_{}C)
B_{}C = {q,r,s}_{}{r,s,t} = {q,r,s,t} ŕ(1) C_{}B = {r,s,t} _{}{q,r,s} ={q,r,s,t}
ŕ(2) From (1) and (2) we conclude that B_{}C =C_{}B 1. _{} B_{}C = {q,r,s}_{}{r,s,t} = {r,s} ŕ(3) C_{}B = {r,s,t}_{} {q,r,s} = {r,s}
ŕ(4) From (3) and (4) we conclude that B_{}C = C_{}B 2. _{} Intersection of sets
is commutative A_{}B = {p,q,r,}_{}{q,r,s} = {p,q,r,s} _{} A_{}(B_{}C) = {p,q,r} _{}{q,r,s,t} ={p,q,r,s,t,} ŕ(5) (A_{}B)_{}C= {p,q,r.s}_{}{r,s,t} = {p,q,r,s,t}
ŕ(6) Since (5) and (6) are same A_{}(B_{}C) = (A_{}B)_{}C 3._{} A_{}B = {p,q,r}_{}{q,r,s} = {q,r} A _{}(B_{}C) ={p,q,r}_{}{r,s} ={r}
ŕ(7) (A_{}B) _{}C = {q,r}_{}{r,s,t} = {r} ŕ(8) Since (7) and (8) are same A_{}(B_{}C) = (A_{}B) _{}C 4._{} Intersection of sets is associative A _{}(B_{}C) = {p,q,r}_{}{r,s} = {p,q,r,s}
ŕ(9) A_{}C = {p,q,r}_{}{r,s,t} = {p,q,r,s,t} (A_{}B) _{} (A_{}C) = {p,q,r,s}_{}{p,q,r,s,t} ={p,q,r,s}
ŕ(10) Since (9) and (10) are same A
_{}(B_{}C) = (A_{}B) _{} (A_{}C) 5._{} A_{} (B_{}C) = {p,q,r,}_{}{q,r,s,t} ={q,r} ŕ(11) (A_{}B) = {p,q,r}_{}{q,r,s} = {q,r} (A_{}C) = {p,q,r}_{}{r,s,t} = {r} (A_{}B)_{}(A_{}C)= {q,r}_{}{r} = {q,r} ŕ(12) Since (11) and (12) are
same A_{} (B_{}C) = (A_{}B)_{}(A_{}C) 6._{} Intersection of sets is distributive over union of sets 

De Morgan’s laws
Prove that
1. (A_{}B)^{1}= A^{1}_{}B^{1} (The complement of union of sets is the
intersection of their complements)
2. (A_{}B)^{1}= A^{1}_{}B^{1}(The complement of the intersection of sets is
the union of their complements)
3.3.1 Example 2 : Let
U = {0,1,2,3,4,5,6,7,8,9}:A = {x: x is a perfect square less than 10}:B
= {x: x is a multiple of 3 less than 10}
Let us write the elements of the sets A and B A = {1,4,9} (other numbers are
not squares of any number) B = {3,6,9} (3 = 3*1,
6=3*2,9=3*3) A^{1 }= UA ( A^{1} contains elements of U which are not the elements of A)^{} = = {0,1,2,3,4,5,6,7,8,9}  {1,4,9}
={0,2,3,5,6,7,8} =========ŕ(1) B^{1}= UB ( B^{1} contains elements of U which are not the elements of B)^{} ={0,1,2,3,4,5,6,7,8,9}  {3,6,9} ={0,1,2,4,5,7,8}
=========ŕ(2) _{}From (1) and (2) we get A^{1}_{}B^{1}= {0,2,3,5,6.7,8}_{}{0,1,2,4,5,7,8} ={0,2,5,7,8} ==================ŕ(3) (A_{}B) = {1,4,9}_{}{3,6,9} = {1,3,4,6,9} _{} (A_{}B)^{1 }= U
(A_{}B) = {0,1,2,3,4,5,6,7,8,9} {1,3,4,6,9} = {0,2,5,7,8} ==ŕ(4) Since (3) and (4) are
same 1. (A_{}B)^{1 }= A^{1}_{}B^{1}^{} From (1) and (2) we get A^{1}_{}B^{1}= {0,2,3,5,6,7,8}_{}{0,1,2,4,5,7,8} ={0,1,2,3,4,5,6,7,8}==============ŕ(5) A_{}B = {1,4,9}_{}{3,6,9}= {9} (A_{}B)^{1}= U – (A_{}B) = {0,1,2,3,4,5,6,7,8,9} {9} ={0,1,2,3,4,5,6,7,8} =======ŕ(6)^{} Since (5) and (6) are same 2. (A_{}B)^{1 }= A^{1}_{}B^{1} 

3.3.2
Relationship between numbers of elements of 2 sets
The number of elements in a set A is
called ‘cardinal number’ of the set and is denoted by n(A).
3.3.2 Example 1 : Let A= {p,q,r,s,t} and
B= {r,s,u,v,w}
n(A) =n(B)=5 A_{}B ={p,q,r,s,t}_{}{r,s,u,v,w}= {p,q,r,s,t,u,v,w} A_{}B ={p,q,r,s,t}_{}{r,s,u,v,w} =(r,s}
_{} n(A_{}B) =8, n(A_{}B) =2 n(A) +n(B) = 5+5 =8+2 = n(A_{}B) +n(A_{}B) These equations can be re written as 1. n(A_{}B)= n(A) +n(B)n(A_{}B) 2. n(A_{}B)= n(A) +n(B)n(A_{}B) 3. When A and B are
disjoint sets n(A_{}B)= n(A) +n(B) ( _{} n(A_{}B)=0 as A_{}B is a null set when A and B are disjoint sets). 

3.3.2 Problem
1: A florist has certain number of garlands. 110 garlands
have Champak flowers in the garlands, 50 garlands have jasmine flowers in the
garlands and 30 garlands have both the flowers in them. Find the total number
of garlands he has.
Solution :
Let A be the set having garlands having Champak, therefore n(A) =110. Let B be the set of garlands having Jasmine, therefore n(B)= 50. A_{}B is the set of garlands which has both these flowers. Therefore n(A_{}B)=30. A_{}B is the set of garlands the florist has. We
know n(A_{}B)= n(A) +n(B)n(A_{}B) = 110+5030 =130 Therefore
the florist has 130 garlands. 

3.3.2 Problem
2: A class has 60 students.
Every one should choose to be in Kabadi team or
hockey team or in both the teams. If 45 students chose to be in Kabadi team and 30 students chose to be in Hockey
team, how many are in both the teams?
Solution :
Let A be the set of students who are in Kabadi
team. Therefore n(A) =45 Let B be the set of students who are in Hockey team. Therefore
n(B) = 30 A_{}B is the set of students who are in both the teams. We are asked to find n(A_{}B). A_{}B is the set of students in the class. It is given that n(A_{}B)=60 We know that n(A_{}B)= n(A) +n(B)n(A_{}B) _{} n(A_{}B)= n(A) +n(B) n(A_{}B)= 45+3060 =15 15 students have taken both Mathematics and Science. 

3.3.2 Problem
3 :A TV viewer ship survey was conducted by an agency.
They conducted a survey on a sample of 1000 families in a place. They found
that 750 families viewed News channel, 400 families viewed sports channel and
300 Families viewed both channels.
Find out
1. How many families viewed News
channel only?
2. How many families viewed Sports
channel only?
3. How many families viewed neither of
the channels?
Solution :
Let U be the set of families who were surveyed Let A be the set of families who viewed News channel. _{} (A) =750 Let B be the set of families who viewed Sports channel. _{} n(B)=400 A_{}B is the set of families who viewed both news and sports
channel. _{} n(A_{}B)=300 Notes: 1. AA_{}B is the set of families who see only News channel and its
numbers is n [AA_{}B]. 2.
B A_{}B is the set of families who see only Sports channel and
its numbers is n [BA_{}B]. 3 A_{}B is the set of families who see either News or Sports
channel. its numbers is n(A_{}B)= n(A)+n(B)n(A_{}B)= 750+400300 = 850 4.
(A_{}B)^{1} is the set of families who neither see News
nor Sports and its numbers is n(A_{}B)^{1} We
have: 1. n [AA_{}B] = n(A) – n(A_{}B) = 750 300 = 450(Only news Channel viewers) 2. n [BA_{}B] = n(B) – n(A_{}B) = 400 300 = 100(Only Sports Channel viewers) 3. n(A_{}B)^{1}= n[U – (A_{}B)] = n(U) – n((A_{}B)) = 1000850 = 150( who do
not view 2 channels) 

3.3 Summary of learning
No 
Points to remember 
1 
(A_{}B)^{1 }= A^{1}_{}B^{1} 
2 
(A_{}B)^{1 }= A^{1}_{}B^{1} 
3 
n(A_{}B)= n(A) +n(B)n(A_{}B) 
4 
n(A_{}B)= n(A) +n(B)n(A_{}B) 
Additional Points:
Let us assume that you have been given the
following question in an examination for matching.
From examination point of view there can only be
three right matches. However, as it can be seen from the figure (Every elements
in set A is paired with every element in set B), we have 12 possible pairs.
Thus we have in all 4*3 = 12 possible pairs.
Observe that (
If A and B are two given sets, the set containing
all the ordered pairs where the first element is taken from A and the second
element taken from B is called ‘Cartesian product’
of two sets. The resulting set is denoted by A_{}B (read as A cross B).
A_{}B = { (x,y)
: x _{} A and y _{} B}
Observe in the above figure that (
Since (a,b)_{}(b,a) A_{}B _{} B_{}A
Note that n(A_{}B) = n(A)*n(B)
Some times when we have an ordered pair, it is
possible to have a relationship between elements of two sets. Some of these
relations are =, <,>, .
A ‘relation’
is a set of ordered pair which explains how elements of one set are related to
elements of other set and the relation is normally denoted by R. Let
A = {Uttar Pradesh, Andhra Pradesh,
B = {
We can have a relationship
called ‘state’s capital is’.
Then the meaningful relationships are {(Orissa,
Note that (Bihar,