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,