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TABLE OF CONTENTS
1.00 INTRO DIFFERENTIAL CALCULUS.
1.10 The limiting value of a function
1.20 The slope of a tangent line
1.30 The continuity of a function.
1.40 Differentiation as excremental notation.
1.50 Techniques of differentiation
1.60 The second derivative
1.70 Turning points: (stationary and inflexion)
Minimum points (stationary and inflexion)
1.80 Business applications
2.00 INTRO INTEGRAL CALCULUS
2.10 Define integration
2.20 Methods of integration
2.30 Solution of definite and indefinite integrals.
2.40 Business applications
3.00 INTRO SET THEORY, PERMUTATIONS AND
COMBINATIONS
1.10 SET Theory: Definition of a set
3.20 Representation / Notation
3.30 Elements of a set, sunsets and supersets,
cardinal number of sets.
3.40 Interrelations of sets: complement,
Intersection, union of sets.
3.60 The venus Diagram (Leonard Ender and John
venn)
3.60 Define permutation and
3.70 List Example of permutations and
combinations.
3.80 permutations with similar / identical objects
conditional permutations.
3.90 Applications of permutations and
combinations.
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INTRO 4.00 BINOMIAL THEOREM
4.10 The meaning of the theorem
4.20 The pascal’s triage: features of the binomial
expansion.
4.30 Use of the combination formula in binomial
expansion.
4.40 Linear approximations and the binomial
expansion.
INTRO 5.00 THE IDEAR OF PROPERBILITY
5.10 Definition and meaning
5.20 Basic rules of probability
5.30 Conditional probability
5.40 Bayes theorem
5.50 Applications to business
INTRO 6.00 LINEAR PROGRAMING.
6.10 The meaning and use of linear programming
6.20 Assumptions of L.P. Model
6.30 Expressing L. P. problems.
6.40 Graphical solution to L. P. problems.
6.50 Simplex method for solving L.P. problems.
6.60 Advantages and disadvantage of graphs and
simplex methods.
INTRO 7.00 TRANSPORTATION MODEL
7.10 Meaning And Use Of The Model
7.20 General features of the model.
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CHAPTER ONE
INTRODUCTION TO CALCULUS
Like most science subjects, mathematics can be divided into several branches,
among which are geometry, Algebra, Arithmetic measurements calculus,
probabilities and statistics and trigonometry. Each branch of subject has several
sub-branches, with each sub-branch serving specific functions. The central theme of
this chapter is the branch called ‘calculus’. Calculus is a wide and complex aspect
of Mathematics, which can be applied in several fields of human endeavour. The
scope of this chapter will however be limited and finance.
There are two main divisions of calculus: differentiation (also call integral
calculus).
Calculus is used for measuring the rate at which a quantity is changing, where
change is represented as slope, gradient or differential coefficient of the graph of
that variable in comparison with another calculus is used to find the gradient of
straight lines and curves. On the other hand, the integral calculus is a term which
describes a method for running up changing quantities. With the integral calculus, it
is possible to determine the areas and volumes of shapes bounded by curves.
Examine the following illustrations fo he descriptions above;
1. Given the Linear equation y = mx + c’
Y = total costs
X = quantity produced/sold and
M and c are constants, can be graphed thus:
Y
Mx
c
x
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The initial part of the graph “C’ is held constant, irrespective of variations in the
value of x, and as such has not rate of change. However, the letter part ‘mx’ increases
as the value of x increases/decreases. Hence, the earlier part ‘c ‘ can be likened to the
fixed part of cost while the later ‘mx’ ca n be described as the variable component of
cost.
II GRADIENT
This simply means rate of change of one quantity measured against another.
Examples of useful gradiets are:
1. rate of change of velocity moved with time, also called accelerations
and;
2. rate of change of quantity against price, also called elasticity
A. Where the relationship between the two variables is linear, that is the graph
of the variables is a straight line, gradient can be found using:
M = vertical distance = y2 – y1
Horizontal distance = X2 – X1.
A gradient is also called a slope, and it must have a sign and size.
Vertical distance
Horizontal distance
Illustration by graph
Y B
Y2 Y-Y
Y A C Slope = m = gradient =y2 – y1 X2 – X1
Xx
B. Where the relationship between the variables is non-linear, this presents a moredifficult problem to solve. The statement Y2-Y1/X2-X1 is still relevant, but will only
give the gradient of the curve between two specific points. In other words, the
gradient of a curve is not the same at all intervals. The process of differentiation
provides a ready means of finding the slopes/ gradients /rates of change of such
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functions and their turning points. Integration, on the other hand, helps to determinethe total amount of costs or revenue between two levels of activity.
111. THE LIMITING VALUE OF FUNCTION
Suppose that y=x2. The graph of y=x2 for values of x from o to 4 is as shown below.y
16
14 C12
10 C
8
6
4 B
2 A
0 1 2 3 4 X
Gradient between A and B = 9 – 4 = 5
3 - 2
Gradient between B and c = 16 – 9 = 54 - 3
NOW EXAMINE THE FOLLOWING GRAPH.
Y B Y2
A Y1
0 Y1 (X1+A) X
From the graph, X1 = X Y1 = X2
X2 = X + a y2 = (x +a)2
Y2 – y1 = (x + a )2 – x2
X2 – X1 = (x + a ) – x
= X2 + 2a x + a2 – x2 = 2ax + a2
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x + a – x a
= 2x + a
Therefore, between A and B, gradient of AB = 2x + a. But as a tends to zero, 2x +
a = 2x. This process of calculating the limiting value of a function is calleddifferentiation.
So that since y = x2
dy = 2x, the limiting value and
dx derivative of y with respect to x.
IV SOLVING LIMITS
The solution of limits can be obtained by avoiding any one of the following
situations:
0/0, 0/z, z/0, &/&, z/&, &/zwhere z = any number (positive or negative)
& = extremely small or large number, also called infinity.
V. METHODS
Some of the methods for solving limits are listed below:
1. Direct substitution
2. Simplification
3. Factorisation
4. Rationalization
5. Differentiation and l’ hospitals Rule.
1. DIRECT SUSTITUTIONS
This should always be attempted before any other method is used. Once the
identified value are avoided, the solution of the limit is good.
EXAMPLES.
(i) line x2 + 2x + 1
x =) a. As x becomes a, x2+ 2n + 1 = a2 + 2a + 1
= ( a+ 1 ) ( a + 1 )
(ii) line x + 22x
x- ) 2
As x ) 2, x + 2 = 2 + 2 = 4 = 12x 2 (2) 4
2. Simplification and fractorisation
Examples
(i) lim x2 – 1 = lim ( x+ 1) (x – 1)
x-)1 x - 1 x-)1 (x –1)
===) lim ( x + 1)
x -) 1
:. As x-)1, x + 1 = 1 + 1 = 2
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(ii) lim x( x + 2 ) (x+ 1 )x-)& x3 – x
==) lim x ( x + 2 )(x+ 1)x-)& x (x2- 1)
but x2 –1 = x2- 12 = ( x – 1 ) ( x + 1 )==) lim x ( x + 2 ) ( x + 1 ) = lim x + 2
x-)& x(x + 1 ) ( x + 1 ).X-)& x- 1
As x-) &, x + 2 = & + 2 = 2 = -2X – 1 & - 1 -1
(iii) Limi 2x3 + 4x2 + xx -) 0 x2 + 5x
Divide all through by the least power of x
==) 2x3 + 4x2 + xx x x
x2 + 5x
x x
= 2x2 + 4x + 1
x + 5
:. As x -) 0, 2x2 + 4x +1 = 2(02) + 4(0) + 1 = 1x + 5 0 + 5 5
3. Rationalization
Lim 2 - x – 3 x 2 + x – 3
x-)7 x2 – 49 2 + x - 3
= Lim 4 +2 x – 3 - 2 x - 3 - (x – 3 )x -) 7 ( x2 – 49) (2 + x – 3
= Lim 4 – x + 3 = -X + 7
X-)7 (x – 7) (x + 7) ( 2 + x – 3 (x – 7) (x+ 7 (2 + x – 3 )
= Lim -1(x –7) = Lim -1
x -) (x-7) (x +7) (2 +x-3 ) x - )7 (x+7) (2+ x – 3 )
= -1 = -1 = -1 = -1
(7+7) (2+ 7 –3) 14 (2 +2) 14 (4) 59
Differentiation ( L’hospital’s Rule)
This should be used where other methods fail, The solution obtained bydifferentiation is the same as that obtained using any other method.
EXAMPLES
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(i) Lim x2 – 1x-) x – 1
Takes x2 – 1 = f (x) and x –1 = g (x)
F’ (x’) = dy/dx = 2x, g’ (x) = 1
:. Lim 2x = 2(1) = 2
(ii) Lim (x – 2) (x2 + 3x –2)
x-)2 x2 – 4
= Lim x3 + 3x2 – 2x – 2x2 – 6x + 4
x-)2 x2 – 4
= Lim x3 –x2 – -8x +4x-)2 x2-4
f’ (x) = 3x2 + 2x – 8, g, (x) = 2x
:. Lim 3x2 + 2x – 8 = 3(22) + 2 (2) – 8x-)2 2x 2(2)
= 12 + 4 – 8 = 8 = 24 4
This solution can equally be obtained thus
Lim (x – 2) (x2+ 3x – 2) = ( x2+ 3x – 2 )x-)2 (x +2 ( x – 2 ) (x + 2)
= 22 + 3 (2) – 2 = 4 + 6 - 22 + 2 4
= 8 = 24
(iii) Lim 2x3 + 4x2 + xX- )0 x2 + 5x
6’(x) = 6x2 + 8x + 1, g’ (x) = 2x + 5
:. Lim 6x2 + 8x + 1 = 6(02 ) + 8 (0) + 1 = 1x-)0 2x + 5 2(0) + 5 5
(VI) THE INCREMENTAL NOTATION.
(THE FIRST FICTION)
1. Take y = x2,Let change in y = dy and change in x = dx
:. Y + dy = (x+ dx)2
y + dy = x2 + 2xdx + dx2
dy = x2 + 2xdx + dx2 – y ( note that y = x2)
dy = x2 + 2xdx + dx2 – x2
dy = 2xdx + dx2
Dividing all through by dx,
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dx = 2xdx + dx2 dx dx dx
dx = 2x + dxdx
As dx-)0, dx = 2x
dx2. Differentiation from the first principle y = x2 + 5x
y + dy = (x + dx)2 + 5(x+dx)
y+dy = x2 + 2xdx + dx2 + 5x + 5dx
y+dy = x2 + 2xdx + dx2 + 5x + 5dx
dy + x2 + 2xdx + dx2+ 5x 5dx – x2 – 5x
dy = 2xdx + dx2 + 5dx
divide through by dx
dy =2xdx + dx2 + 5dx
dx dx dx dx
dy = 2x + dx + 5
dx
As dx-)0, dy = 2x + 5
dx
dy = 2x + 5dx
3. Given y = 3x2 – 2, differentiate y with respect to x from the first principle.
Y = 3x2 – 2
Y + dy = 3(x + dx)2 – 2
Y + dy = 3(x2 +2dxdx + dx2) –2
Y + dy = 3x2 + 6xdx + 3dx2 –2
Y + dy = 3x2 + 6xdx + 3dx2 – 2 – 3x2 + 2
dy = 6xdx + 3dx2
dx = 6xdx + 3dx2 dx dx dx
dy = 6x + 3dx.dx
As dx-)0, dy = 6x
dx
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4. Differentiate from the first principle y = 2x3 + 4x2 + 6x + 3
y + dy = 2(x+dx)3 + 4 ( x + dx )2 + 6 ( x + dx ) + 3
y + dy + 2(x3 + 2x2dx + xdx2 + x2dx + 2xdx2 + dx3) + 4(x2 + 2xdx + dx2) + 6x + 6dx
+ 3.
Y + dy = 2(x3 + 3x2dx + 3xdx2 + dx3) + 4x2 + 8xdx2 + 6x + 6dx + 3
Y + dy = 2x3 + 6x2 dx + 6xdx2 + 2dx3 + 4x2 + 8xdx + 4dx2 +6x + 6dx+3
dy = 2x3 + 6x2 dx + 6xdx2 + 2dx3 + 4x2 + 8xdx + 4dx2 +6x + 6dx+3
-2x3 – 4x2 – 6x –3
dy = 6x2 dx + 6xdx2 + 2dx3 + 8xdx + 4dx2
dy = 6x2 dx + 6xdx2 + 2dx3 + 8xdx + 4dx2 + 6dx
dx dx dx dx dx dx dx
dy = 6x2+ 6xdx + 2dx2 + 8x + 4dx + 6
dx
As dx -)0, dy = 6x2 + 8x + 6
5. Once again, consider y = x2, dy 2x from this,
dxwe can infer that fir every y = axn,dy/dx = anx n-1
Therefore, given y = axn,dy/dx = anxn-1
The blocked formula is considered the general principle of differentiation.
Vii TECHNIQUIES OF DIFFERENTIATIONA. Monomial and polynomial
In general, for every y = axn, dy = anxn-1
dx
EXAMPLES
1. Find dy/dx if y = mx,
dy/dx = m (mx1 x X1-1).
2. Differentiate with respect to x,
y = 6x2
dy/dx = 12x (6x2xX2-1).
Where the function is a sum, the differentiate coefficient is the sum of the
differentiate coefficients of the separate parts.
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3. Given y = 2x3 + 4x2 + 6x
dy/dx = 6x2 + 8x +6
To differentiate a whole number, say 5, attack x as follows,
4. Given that y = 3x2 + 7x + 5
y = 3x2 + 7x + 5x0
dy/dx = 6x + 7 + 0
= 6x + 7 =
B PRODUCT RULE
Let u and v represent functions of x and y, such that y = uv, by implication
dy of uv = u dv + v du
dx dx dx
Examples
1. Find dy/dx if y = (2x + 3) (3x2 + 2)
Take 2x + 3 = 3 = u and 3x2 + 2 = v
Since u = 2x + 3 dx2 3x22 + 2
du = 2 dv = 6xdx dx
From dy u dv/dx + v dv/dx
dxdy = (3x2 + 2)2 + ( 2x + 3) 6x ( note: a+b = b + a).
= 6x2 + 4 + 12x2 + 18x
=18x2 + 18x + 4
= 2(9x2 + 9x + 2)
2. Find dy/dx in y = (3x – 1) (4x +6)
Take u = 3x – 1 and v = 4x + 6
du = 3 and dv = 4dx dn
:. dy = (3x – 1)4 + (4x + 6)3dx
=12x – 4 + 12x + 18
= 24x +14
= 2(12x + 7)
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3. If y = (x2 + 4 ) ( 6x1/2 + 3 )
du = 2x and dv =3x-1/2
dx dx
dy = (x2 + 4) 3x-1/2 + (6x1/2 + 3) 2x.
= 3x3/2 + 12x-1/2 + 12x3/2 + 6x
= 15x3/2 + 6x + 12x-1/2
4. Find dy/dx given that y = (3x2 – 2x + 1) (4x +5)
Take 3x2 – 2x + 1 = u and 4x +5 = v
dy = v du = (4x +5) (6x – 2) + (3x2 –2x +1)4dx dx
= 24x2 – 8x +30x – 10 + 12x2 – 8x +4= 36x2 + 14x –6.
C QUOTIENT RULE
Let u represent the function of x which is the numerator and v represent that which
is the divisor.
Consequently, y = u
V
dy = v du = u dv
dx dx dx
v2
Examples
1. differentiate y = x2 - 1x3 + 1
Take x2 – 1 = u and x3 + 1 = v
du = 2x dv = 3x2
dx dx
dy = v du - u dv = (x3 + 1) 2x – (x2 – 1) (3x2)
dx dx dx (x3 + 1) (x3 + 1)v2
= 2xn + 2x – 3x4 + 3x2 = -x4 + 3x2 +2x
x6 + x3 + x3 + 1 x6 + 2x3 +1
2. Find dy if y = x - 1
dx x +1
Take u = x – 1 and v = x + 1
d du 1, dv = 1
dx dx
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dy = (x + 1)1 - ( x – 1)1 = x +1 – x + 1 = 2
dx (x + 1)2 x2 + 2x + 1 x2 + 2x + 1
3. If y = 4x3 + 2, find dy
x2 dx
Take u = 4x3+ 2 and v = x2
du = 12x2, dv = 2x
dx dx
dy = x2 (12x2) – (4x3 + 2) 2x = 12x4 – 8x4 – 4x
dx (x2)2 x4
4x2 – 4x = 4x (4x3 -2) = 4(x3 – 1)
x4 x(x 3) x3
4. Given y = 3 – 2x find dy
3 + 2x dx
Take u = 3 – 2x and v = 3 + 2x
du = -2, dv = 2dx dx
dy = ( 3 + 2x ) –2 – (3 – 2x)2 = - 6- 4x – 6 + 4x 4xdx (3 + 2x)2 9 + 12x + 4x2
= -12(3 + 2x)2.
CHAPTER TWO
SET THEORY, PERCUTATIONS
AND COMBINATIONS
I. SET THEORY
A. DEFINITION
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“Is a well-defined list, collection, or class of objects. The objects in sets can be
anything: numbers, people, letters, rivers, c. t. c. these objects are called the
elements or members of the set”.
……Mayo associates limited & BPP publishing limited.
“A set is a collection of things. The things have something in numbers. The
example, a set of stamps, a set of chairs, a set of numbers. The things in a set are
called its members or elements”.
…Duncan And Christine Graham, Gose Mathematics.
“ A set is simply collection of things or people, called elements.
Capital letters such as A, B, C, are used to named such collections.
…. Walter van stigh, success in mathematics.
“Mathematics call any well-defined class a set, and by well-defined we mean that
we must be able to decide definitely whether any one object does or does not belong
to that set”.
….F. G. J. Norton,. Hardwood Clarke’s ordinary level mathematics
The above definitions set out clearly the subject matter of this chapter. The
language of sets has been one form in the flesh for most students learning
mathematics or mathematics – related courses. While, the key words in the
definitions above can be cited as.
‘well-defined’ ‘collection’
‘elements or members’
B. NOTATION
Notation means representation of sets. How are sets written (represented)? Which
rule(s) must be observed in drawing sets? This
Parts provide answer to the other possible questions.
=(x2 +4)3 (33x4 + 36x2)
Generally, sets are donated by capital letters and written in
= (x2 + 4)3 (33x4 + 36x2)
inverted brackets, the elements are not arranged in any particular
= 3x2(x2 + 4)3 ( 11x2 + 12). Order.
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Examine the following expert opinions on set notation:
“A set is usually named XVIII. APPLICATION PROBLEMS or labeled with a
capital letter. Curly brackets or braces () often replace the words the set of ‘. And
when members are listed the order does not matter. Commas must be placed
between members”
…Duncan and Christine graham ditto
“Capital letters such as A, B, C, are used to name such collections
by: a. describing the set in words;
i. Nis the set of al integers
ii. A is the set of all prime numbers
iii. q is the set of all past presidents of America.
B. Listed or tabulating all the elements of a set; which is only practicable if the set
is relatively small, The brackets () are used to enclose the listed elements of a set or
its description by means of a property of its elements”.
….Walter van stigt, success in mathematics.
“ sets are usually donated by capital letters
A, B, X, Y, S,……. The elements in sets are usually donated by small letters a, b, x,
y, s, ….. if A is the set of numbers 1, 2, 7, and 10, then we write A = (1, 3, 7, 10)”.
….. Mayo and BPP. Do.
From the above descriptions, a set of all positive even between o and 10 inclusive
will be written thus:
A = (2, 4, 6, 8, 10)”.
Note also that sets can be given as inequalities. Imagine a manufacturing process,
which requires its operatives to produce a minimum of 5 units but not more than 10
units of product daily. The set can be written thus:
P5 p 10
That is 5 p, otherwise written as p 5
And p 10
P = (5, 6, 7, 8, 9, 10).
In order to be able to display or decide sets which are given as inequalities, a fairly
good understanding of the following signs is essential:
‘less than’ ‘greater than’
‘less than/equal to’ ‘ greater than /equal to’
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# ‘not equal to’
c. TECHNICALITIES IN NOTATION
1. The use of…………
a. Continuing a pattern to infinity.
Suppose x = (a, 2a, 3a, 4a, ……) the after the 4th member the pattern
continues indefinitely (say to infinity)
b. continuing a pattern to a limit
suppose y= (a, b, c, d,…..z) The …suggests that the pattern continues until z. thus,
the last term of the series is z.
2. The cardinal number of set.
This is simply the number of elements contained in the set Hence, if p = (5, 6, 7, 8,
9, 10,), n(f) = 6
And if y = (a, b, c, d, …..z), n(z) = 26
3.The universal set (u or e)
This is the which consists of everything, all members, all members, of a set under
examination. U or E donates it.
Suppose set A = ( all white lorries)
Set B = ( all blue lorries)
Set C = ( lories of other cal curs)
U = (all lories)
Similarly, Let X = (all primary school pupils)
Y = (all secondary school pupils)
Z = (all undergraduates)
T = ( all postgraduates students)
U = (all students).
4. SUBSETS
A subset can be defined as a smaller set within a bigger set.
It is thus a set inside a set.
A set of all even numbers is a subset of the set of all integers.
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Given that A = (a, b, c, …..z) and B = (all consonants),
C = (all vowels), then Band are subsets of A, written thus:
ECA
CCA
The symbol ‘c’ means is a subset of A = all sets are subsets of their universal sets.
5. The complement of a set.
“ All the elements of the universal set which are not members a set A themselves
from a set called the complement of A, written A”.
“These set of all element s of the universal set which are not elements of set A is
called the complement of A. I is usually symbolized by a dash (‘) after the
appropriate letter”
W. V. STIGT.
From the above vices, the complement of a set is found by comparing that set with
the universal set.
Suppose set A = (Tuesday, Wednesday, Friday)
B = ( Thursday, Sunday)
U = (Monday, Tuesday,….Sunday)
Thus A = (Sunday, Monday, Thursday, Saturday)
B = (Monday, Tuesday, Wednesday, Friday, Saturday)
Similarly, given that A = (2, 4, 6, 8, 10)
B = (1, 3, 5, 7, 9,)
And U = ( 1, 2, 3, ….10)
Then A’ = (1, 3, 5, 7, 9)
B’ = (2, 4, 6, 8, 10)
6. The Union sets
A List of all the elements contained in two or more given sets is the union of the
given sets.
For example, let A = (chemistry, physics, Biology)
B = (Economics, physics, commerce)
The set of union B, written as A V B is
A V B = (Chemistry, physics, Biology, commerce, economics)
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From the above, you would see that all the elements in A and B were organized and
presented as AVB.
You will also have noticed that ‘physics’ which is element of A and B, has been
written just once. This is because, by rule, when compiling the union of sets,
elements common to two or more of the sets must be written only once.
Similarly, given A = (1, 2, 4, 8) and B = (1, 3, 5, 7)
AVB = (1, 2, 3, 4, 5, 6, 7, 8)
Thus if an element can belong to the set P or to the set Q, it belongs to the union of
P and Q. This is the set of all the elements of P or Q or of P and Q.
7. THE INTERSECTION OF SETS.
Two set are said to intersect when each has element common to the other. The
intersection of two or more sets is, A n B, for sets A and B
Alternatively, if A and B are two given sets, An B = ( x/x EA, XEB).
The above needs, / intersect B is equal to x, such that x is a member of A and X is a
member of B.
Given A = (1, 2, 3, 4) and = (3, 4, 5, 6)
A n B = (3, 4)
8. EQUALITY OF SETS
Two sets are equal when they contain the same set of elements or members. The
manner (order) of arrangement of the elements does not matter.
For instance, if A = (a, c, c, z) and
B = (z, c, c, a)
A = B
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D. THE VENN DIAGRAM
This is a diagrammatic representation of the interrelationship between two or more
sets and the universal set,
The diagram was invented by Leonard Enler and simplified and improved by John
Venn. It is thus named after him.
The universal set (E or LL) is represented with a rectangle (or square) while others
are represented with circles.
Now examine the following illustrations.
(i) A = (Ade, Bayo, Sola, Kola, Bode)
Ade
Bayo Kola
Sola
Bode
(ii) Given P = (1, 2, 3, 4, 5, 6, 7, 8)
U = (1, 2, 3,………..10).
9 1 3 2
6 4 5
7 8 10
(iii) suppose x = (1, 2, 4, 8)
Y = (3, 4, 6)
Z = (1, 4, 8)
X n z = (4), y n z = (4)
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X Y
2 4 3
1 8 6
(iv) If the universal set for x, y and z above is u = (1, 2, 3, 4, 5,…..10)
The Venn becomes
X 2 -- 3 Y
1 2 4 --
--
Z
(V) For sets x and z
X 2 1
2 4 8
The diagram below further summarizes the interelationship of sets, using Venn
diagrams.
S/N VENN DIAGRAMS NOTATION INTERRETATION
1 . U
E or U Universal set
2. n (x) as The set of elements
Subset of u common to x and y
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3.
x Y
X n Y the set of elements common to x and
Y.
4. X Y x -y The difference of sets x and y.
Also called x only.
5.
Y x - y The complement
of sets x .
6. x u y Set of all elements
in x and y or
of x or y.
7.
X’ The complement of x
X
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8. A B B C A B is a subset of A
9.
P n Q = ( ) P and Q are
P Q Disjoint sets.
10.
A B AURVC The union of three
Sets A, B, and C. a
All elements in A,
B, and C.
C
11.
X Y x n y n z The union of three
Common to x, y
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And z.
Z
12.
X Y (x n y n z)’ The complement
Of x n y n z. The
Elements absent
In x n y n z but
Z present in the
Universal set.
13.
X Y (x u y u z) The complement
Of x u y u z. The set of
elements absent in x u y
u z but present the
universal set.
Z
14.X Y Y n Z The intersection of Y and
z. The set of elements
common to Y and z.
Z
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15.
X Y (XUY) The intersection of
The union of X and Y with Z.
The set of all elements
E. SPECIAL LAWS
THE ALGEBRA OF SETS HAS CERTAIN SPECIAL LAWS:
I. COMMUTATIVE
Generally, in binary operations, if the order of combination of, elements does not
affect results, the operation is referred to as commutative.
In a + b = b + a, addition is commutative but in a – b = b – a, subtraction is shown
no to be commutative.
For sets, A U B = B U A and
A n B = B n A.
II. ASOCIATIVE.
We can combine any three element in two ways. Suppose we intend to add a, b, and
c. Then, we either say ( a + b ) + c or a + (b + c).
Where the order of combination is immaterial, we refer to the operation as
associative.
For sets, (A U B) V C = A U ( B U C)
(A N B ) N C = A n (A n c)
III DISTRIBUTIVE
Suppose a, b and c are three integers, then a (b + c) = ab + ac. This shows that
multiplication is distributive over addition. But (b +c) :-a is not b :-a + c :-a.
Therefore division is not distributive over addition.
For sets, AU (B n C) = (A U B) n (A U C)
A n (B u c) = ( A n B) Y (A n C)
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IV IDENTITY
Identity laws are general laws which affect sets,
These are:
(i) For empty sets ( null sets)
1. AU ( ) = A
2. An ( ) = ( )
3. ( ) UE = E
4. ( ) n E = ( )
5. n (A U B) = n (A) + n( B) – n(A NB).
ii For universal sets ( E or U)
1. A U E = E
2. A n E = A
3. A C E.
4. A’ U E = E
5. A’ n E = A’
6. E’ = ( )
7. ( )’ = E
8. (A’) = A
9. (E’) = E.
V DE –MORGAN’S
1. A’ n B’ = (A U B)’
2. A’ U B’ = (A n B)’
F. WORKED EXAMPLES
(a) (i) Define a set
(ii) let set s = (3,2,5,8, 16, 13) and T = ( 1, 20, 19, 5, 7, 8)
(iii) Let s = (5, 3, 1) and T = (2)
Determine s and T
(iv) Let set A = (x/x2 = 4, x is odd)
What sort of set is A
Use venn diagrams where applicable.
SOLUTION
(a) (i) A set a well – defined list, collection of objects.
(ii) S = (3, 2, 5, 8, 16 13), T = (1, 2, 0, 19, 5, 7, 8)
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S n T = (5, 8)
S T
3 2 5 17
16 13 8 20
19
(iv) A = ( x/x2 = 4, x is odd)
For x2 = 4, x = + = +2
4
X cannot be odd.
:. A = ( ), an empty set.
A survey of a small town of 20000 households provided the following data.
60% have refrigerator;
35% have telephone;
15% have colour television;
18% have telephone and colour television;
12% have all three.
Using a Venn diagram,
a. How many of these households have a telephone?
b. How many have only refrigerator?
c. How many only colour television set.
d. How may have neither a telephone, nor a refrigerator, nor colour
television set?
e. What percentage of the household have only two of the three equipments?
SOLUTION
100% Of The Households = 20,000
60% = 60 X 20,000 = 12,000
100
35% = 7,000
15% = 3,000
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18% = 3,600
12% = 2,400
10% = 2,000
8% = 1,000
HINT:
a. Represent each set with a letter, A, B, C, X, Y,Z.
b. Since It Is Already Established that each set has one thing or the other in
common with the next, represent each set with a circle, showing clearly their
meeting points.
c. Fill the sets, starting from the mid-point common to all.
d. Solve as required.
R T
E C F let refrigerator = R
“ Telephone = T
A “ colour T. Y = C.
C D
G
C
At point a, all three meet = 8% = 1600 households
,, ,, b, R and T meet = 13%= 10% = 200 households
,, ,, c, R and c meet = 10% - 8% = 2% = 400 households
Recall that: a + b = 18%
a + c = 10%
a + d = 12%
Similarly, a + b + d + f = n® = 60% = 12% = 12,000 households
:. C = 12,000 – (1600 + 2000 +400) = 8,00 households
Also, a + b + d + f = n(T) = 35% = 35% = 7,000 households
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:. F = 7,000 – (1600 + 2000 + 800) = 2, 600 households
And a + c + d + g = n(c) = 15% = 3,000 households
:. G = 3,000 – (1600 + 400 + 800) = 200 households
THE RE-DRAWN VENN IS THUS:
R 2u 2b T From this redraw venn, we
2nc can now answer all the
2ub questions.
4n f2n
2cn
C
(a) n (Telephone) only = 2600 households
(b) n ( Refrigerator) only =8000 households
(c ) n (colour T.V) only = 200 households
(d) n ( neither T nor R no c)
= 2000/0 – n (either T or R or C)
= 200,60 – (8000 + 2000 + 1600 + 400 + 2600 +800 + 200)
= 20,000 – 15, 600 = 4, 400 households.
THE VENN CAN ALSO SHOWN IN PRECENTAGES
R T
40% 10% 13%
8%
2% 4% 22%
19% C
d. n ( % having only two) = 10% + 25 + 4% = 16%.
3. In a certain examination, 72 candidates offered Mathematics, 64 English and
62 French. 18 offered both Mathematics and English, 24 Mathematics and
French and 20 English and French. 8 offered all three subjects . How many
candidates were there for the examination?
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SOLUTION
A 34
10 Let mathematics = M
8 82 English = E
E 38 26 F French = F
16
n (M n E n F) =8 n ( M n F) = 24
n (m n E) = 18 n (E n F) = 20
n( M ) = 72 n (E) = 64
n (F) = 62
n (m n F) only = 24 – 8 = 16
n ( E n F) only = 20 – 8 = 12
n (M n E) only = 18 – 8 =10
n (M) only = n (m) – n(m n E) only + n ( m n E n F)
+ n (m n F) only
n ( m) only = 72 – (10 + 8 + 16)
= 72 – 34 = 38
Similarly, for E,
N (E) only = n (E) – ( n (m n E) only + n (m n E n F) + n ( E n F) only
= 64 – (10 + 8 12)
= 64 – 30 = 34
And for F,
N ( F) only = n (F) – (n (m n F) only + n ( n F) only
+ n ( m n E n F )
= 62 – (16 + 12 + 8)
= 62 – 36 = 26
Total number of candidates for the examination
= 38 + 10 + 34 + 16 + 8 + 12 + 26
= 144 candidates.
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II. PERMUTATIONS
A.DEFINITION
“ A permutation is an ordered arrangement of items. Thus AB is a different
permutation to BA even though the individual two items A and B are the same, they
are in a different order” Terry Lucy, quantitative Techniques.
“ permutation is a technical word for arrangement. A permutation is a set of items,
selected from a large collection of items, in which the order of arrangement is
significant.
…. M. A. Lawal, Business Mathematics volume Two.
“ A permutation of a n different objects with taken or at a time is an arrangement of
out of the n objects with donated by n pr p (n,r) or pn, r and is given by
npr = n(n – 1) (n-2)….(n – r +1) = n’
(n – r)’
Suppose we create boxes for the first three positions
Thus:
1st 2nd 3rd
Any one of any one of any one of
8 1 6
:. Possible ways = 8 x 7 x 6 = 336 ways.
B. THE SPECIAL.
(X’ CXM)
a product such as 5 x 4 = 3 x 2 x 1 is known as factorization written as 5’.
There were 3 x 2 x 1 = 3.
6 x 5 x 4 x 3 x 2 x 1 = 6;
Thus, n’ = n x (n – 2 ) x ( n – 3) x….
From the differentiation given by Lucy special, in America series statistics,
N m = n:
(n – r)
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where in is the total number items available and is the number required or space
available.
In our first example, 3 and 4 players and need also to our n = 4 and r = 4
:. The number of arrangement is
44 = 41 = 41 = 41
(4 – 4)’ 0’ 1
note that 11 = 1 and 01 = 1
:. Nor where n = r = n’
:. 4p4 = 41 = 24 wives.
Similarly, in the second example, n = 8 and r = 3
:. The number of arrangement = 8p3 = 81
(8- 3)
= 81 = 40320 = 336 ways
5’ 128
Examples.
1. How many different numbers can be formed using the digits 2, 2, 3,
3, 3, if the numbers must all be of 5 digits?
The result is 1 , since 2 appear twice and 3 appears
2 3’ twice.
:. Number of 5 digit numbers = 120 = 10
12
2. An airline twice compare is to arrange 3 footballers, 3 boxers and 3 sprinters
on a Lufthansa flight to the International Olympic committee headquarters in
Switzerland. In how many ways can they be arranged if:
(a) each of the sportsmen must remain with their groups and;
(b) only the sprinters must sit together?
(a) Arranging each group together,
000 000 000
F B S
3’ X 3’ X 3’
6 X 6 X 6 = 216 WAYS.
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But note that any one of the three groups can be arranged first = 3’
:. Number of possible arrangements = 216 x 3’ = 1296 ways.
(b) Arranging only sprinters together, this puts footballers and boxers in any
order.
That is 6’ sprinters together = 3’
Call also that either sprinters or a combination of footballers and boxers can be
arranged first. This is also 2’
:. The result is 6’ x 3’ x 2 = 8640.
D. CONDITIONAL PERMULATIIONS
The second example above has already produced this topic.
Sometimes, certain special conditions may be attached to calculating
permutation.
4. A food center offers a choice of 3 starters, 4 main courses and 2 sweets.
How many different meals are available ?
The result is 3 x 4 x 2 = 24 meals.
5. SILOT LTD has 4 training officers and 2 sections requiring their services. In
how many ways may the 4 officers be assigned to the 2 section?
Let the officers be A, B, C, D, and department p, Q,.
P can have any one of remaining 3
= 4 x3 = 12 ways.
Alternatively, number available n = 4
Number required r = 2
:. Possible assignments = npr = 4p2 = 4
(4 – 2)’
= 4 = 24 =12ways
2’ 2
5. In how many ways on the letters of the word POLYTHECHNIC be
arranged?
n= 11, r = 11
:. npr = 11 p11 = 111 = 39,916,800 ways
C. PERMUTATION WITH SIMILAR /INDENTICAL OBJECTS
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Where items to arranged are not completely different, the number of
permutation will be reduced.
Suppose we want to find out the number of the ways possible to arrange
the letters of the word ONDO. The letter appears twice and you cannot
possible distinguish one from the other. If all letters were different, the result
will be 4 p4 = 4 = 24 ways.
But since appears twice, has 21 and our result becomes
4’ = 24 = 12ways
2 2
Examples
1. In how many ways may the letters of the word SCHOOL be arranged if:
(a) The two o’s must not come together and
(b) The two o’s must remain together.
(a) By excluding the o’s in school, the remaining letters S C H L can be
arranged in 4’ ways.
The first o can occupy any one of the remaining four positions vacant = 4.
Also, recall that it is not possible to distinguish between the two o’s = 2
Therefore, the number of arrangements, with the first o above can occupy.
4’ x 5 x 4 = 24 x 5 x 4 = 240 ways.
2’ 2
Examine this diagrams for the spaces which the first o above can occupy.
? s c H L
5 7 C H L
S C ? H L
S C H ? L
S C H L ?
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(B) if the two os must remain together is only change in the result is that whichever
space will be occupied by the first o is automatically for both. That is, 5 spaces
:. The result is 4’ x 5 = 60 ways
2'
2. Examine the word O F F I C E R. In ho w many ways may the letters be arranged
given that:
(a)The F F should be separated;
(b) The F F should remain undetached.
(a) 0 1 c e r = 5’ ways
1st F 1 2 3 4 5 6 =6 places
1 1 1 1 1
2nd
F 1 2 3 4 5 = 5 places
For 2nd F, remember that 1st F would have occupied one out of the 6 places.
Also, the FF ar not distinguishable = 2’
Therefore, number of arrangements
= 5’ x 6 x 5 = 1800 ways
2’
(b) 0 1 C E R =5’ WAYS
F F 1 2 3 4 5 6 =6WAYS
Therefore, the result is 5’ x 6u = 360 ways
2’
III. COMBINATIONS
DEFINITION
“ There will be occasions when selections will be made where the order does not
matter meaning that a, b, is the same as b, a. This is known as a combination”.
…. T. Lucy, quantitative Techniques.
“ In combination, we are concerned with the number of different grouping of
objects that can occur without regard to their order”.
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…. c..Clidi, Business mathematics, volume Two.
“.. is a technical word use in mean session. A combination is a set of items, selected
from a large section of items, regardless of the order in which they are selection.
…. M.A. Lawal, Business Mathematics, volume Two.