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The Educational Publisher

Columbus, 2015

ALGEBRAIC PROBLEMS AND EXERCISES

FOR HIGH SCHOOL

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

Algebraic problems and exercises for high school

1

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

ALGEBRAIC PROBLEMS AND EXERCISES FOR HIGH SCHOOL

Sets, sets operations Relations, functions Aspects of combinatorics

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

2

First imprint: Algebra n exerciii i probleme pentru liceu (in Romanian),

Cartier Publishing House, Kishinev, Moldova, 2000

Translation to English by Ana Maria Buzoianu (AdSumus Cultural and Scientific Society)

The Educational Publisher

Zip Publishing

1313 Chesapeake Ave.

Columbus, Ohio 43212, USA

Email: [email protected]

AdSumus Cultural and Scientific Society

Editura Ferestre

13 Cantemir str.

410473 Oradea, Romania

Email: [email protected]

ISBN 978-1-59973-342-5

Algebraic problems and exercises for high school

3

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

ALGEBRAIC PROBLEMS

AND EXERCISES

FOR HIGH SCHOOL

Sets, sets operations

Relations, functions

Aspects of combinatorics

The Educational Publisher

Columbus, 2015

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

4

Ion Goian, Raisa Grigor, Vasile Marin, Florentin Smarandache,

and the Publisher, 2015.

Algebraic problems and exercises for high school

5

Foreword

In this book, you will find algebraic exercises and problems,

grouped by chapters, intended for higher grades in high schools or

middle schools of general education. Its purpose is to facilitate

training in mathematics for students in all high school categories,

but can be equally helpful in a standalone work. The book can also

be used as an extracurricular source, as the reader shall find

enclosed important theorems and formulas, standard definitions

and notions that are not always included in school textbooks.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

6

Contents

Foreword / 5

Notations / 7

1. Sets. Operations with sets / 9

1.1. Definitions and notations / 9

1.2. Solved exercises / 16

1.3. Suggested exercises / 32

2. Relations, functions / 43

2.1. Definitions and notations / 43

2.2. Solved exercises / 59

2.3. Suggested exercises / 86

3. Elements of combinatorics / 101

3.1. Permutations. Arrangements. Combinations.

Newtons binomial / 101

3.2. Solved problems / 105

3.3. Suggested exercises / 125

Answers / 137

References / 144

Algebraic problems and exercises for high school

7

Notations

= equality;

inequality;

belongs to;

doesnt belong to;

subset;

superset;

union;

intersection;

empty set;

(or) disjunction;

(and) conjunction;

= {0, 1, 2, 3, 4, }

= { ,2,1, 0, 1, 2, }

p equivalent with ; natural numbers

integer numbers set;

= {

|, , 0} rational numbers set;

real numbers set;

= { + |, , 2 = 1} complex numbers set;

+ = { | > 0}, {,, };

= { | < 0}, {,,};

= { | 0} = {0}, {, , ,, };

||

[]

{}

absolute value of ; integer part of x R; fractional part of , 0 {} < 1;

(, ) pair with the first element

and the second element

(also called "ordered pair");

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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(, , ) triplet with corresponding elements

, , ;

= {, | , } Cartesian product of set

and set ;

= {, , | , , }

Cartesian product of sets , , ;

universal set;

() = {| } set of parts (subsets) of set ;

= () ( )

equality of sets and ;

() ( )

is included in ;

= { | } union of sets and ;

= { | } intersection of sets and ;

= { | } difference between sets and ;

= ( ) ( ) symmetrical difference;

() = = complement of set relative to ;

relation defined on sets and ;

: function (application) defined on

with values in ;

() domain of definition of function :

() domain of values of function .

Algebraic problems and exercises for high school

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1. Sets. Operations with sets

1.1. Definitions and notations

It is difficult to give an account of the axiomatic theory of sets at

an elementary level, which is why, intuitively, we shall define a set as a

collection of objects, named elements or points of the set. A set is

considered defined if its elements are given or if a property shared by

all of its elements is given, a property that distinguishes them from the

elements of another set. Henceforth, we shall assign capital letters to

designate sets: ,, , , , , , and small letters for elements in sets:

, , , , , , etc.

If is an element of the set , we will write and we will

read " belongs to " or " is an element of ". To express that is not

an element of the set , we will write and we will read " does

not belong to ".

Among sets, we allow the existence of a particular set, noted as

, called the empty set and containing no element.

The set that contains a sole element will be noted with {}.

More generally, the set that doesnt contain other elements except the

elements 1, 2, , will be noted by {1, 2, , }.

If is a set, and all of its elements have the quality , then we

will write = {| verifies } or = {|()} and we will read: "

consists of only those elements that display the property (for which

the predicate ( ) is true)."

We shall use the following notations:

= {0, 1, 2, 3, } the natural numbers set;

= {0, 1, 2, 3, } the natural numbers set without zero;

= { ,2,1, 0, 1, 2, } the integer numbers set;

= {1,2,3 } the integer numbers set without zero;

= {

|, , } the rational numbers set;

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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= the rational numbers set without zero;

= the real numbers set;

= the real numbers set without zero;

+ = { | 0}; + = { | > 0};

= { + |, , 2 = 1} = the complex numbers set;

= the complex numbers set without zero;

{1, 2, , } = 1, ;

() = { | } = the set of all integer divisors of number

;

() = || = the number of the elements of finite set A.

Note. We will consider that the reader is accustomed to the

symbols of Logic: conjunction (and), disjunction (or),

implication , existential quantification () and universal quantification

().

Let and be two sets. If all the elements of the set are also

elements of the set , we then say that is included in , or that is

a part of , or that is a subset of the set and we write . So

()( ).

The properties of inclusion a. () , (reflexivity);

b. ( ) (transitivity);

c. (), .

If is not part of the set , then we write , that is,

()( ).

We will say that the set is equal to the set , in short = ,

if they have exactly the same elements, that is

= ( ).

The properties of equality Irrespective of what the sets , and may be, we have:

a. = (reflexivity);

b. ( = ) ( = ) (symmetry);

Algebraic problems and exercises for high school

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c. ( = = ) ( = ) (transitivity).

With () we will note the set of all parts of set , in short

() .

Obviously, , ().

The universal set, the set that contains all the sets examined

further, which has the containing elements nature one and the same,

will be denoted by .

Operations with sets Let and be two sets, , ().

1. Intersection.

= { | },

i.e.

( ), (1)

( ), (1)

2. Union.

= { | },

i.e.

( ), (2)

( ), (2)

3. Difference.

= { | },

i.e.

( ), (3)

( ), (3)

4. The complement of a set. Let (). The difference

is a subset of , denoted () and called the complement of

relative to , that is

() = = { | }.

In other words,

() , (4)

() . (4)

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Properties of operations with sets = , = (idempotent laws)

= , = (commutative laws)

( ) = ( );

( ) = ( ) (associativity laws)

( ) = ( ) ( );

( ) = ( ) ( ) (distributive laws)

( ) = ;

( ) = (absorption laws)

( ) = () ();

( ) = () () (Morgans laws)

Two "privileged" sets of are and .

For any (), we have:

,

= , = , () = ,

= , = , () = ,

() = , () = ,

(()) = (principle of reciprocity).

Subsequently, we will use the notation () = .

5. Symmetric difference.

= ( ) ( ).

Properties. Irrespective of what the sets , and are, we

have:

a. = ;

b. = (commutativity);

c. = = ;

d. () = ;

e. () = () (associativity);

f. () = ( )( );

g. = ( ) ( ).

Algebraic problems and exercises for high school

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6. Cartesian product. Let and be two objects. The set

{{}, {, }}, whose elements are the sets {} and {, }, is called an

ordered pair (or an ordered couple) with the first component and

the second component and is denoted as (, ). Having three objects

, and , we write (, , ) = ((, ), ) and we name it an ordered

triplet.

Generally, having objects 1, 2, , we denote

(1, 2, , ) = (((1, 2)3))

and we name it an ordered system of elements (or a cortege of

length ).

We have

(1, 2, , ) = (1, 2, , )

(1 = 1 2 = 2 = ).

Let , (). The set

= {(, )| }

is called a Cartesian product of sets and . Obviously, we can define

= {(, , )| }.

More generally, the Cartesian product of the sets 1, 2, ,

1 2 = {(1, 2, , )| , = 1, }.

For = = = 1 = 2 = . . . = , we have

2, 3, . ori

For example, 3 = {(, , )|, , }. The subset

= {(, )| } 2

is called the diagonal of set 2.

Examples.

1. Let = {1,2} and = {1, 2, 3}. Then

= {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

and

= {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)}.

We notice that .

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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2. The Cartesian product 2 = can be geometrically

represented as the set of all the points from a plane to wich a

rectangular systems of coordinates has been assigned,

associating to each element (, ) 2 a point (, ) from the plane

of the abscissa and the ordinate .

Let = [2; 3] and = [1; 5], (, ). Then can be

geometrically represented as the hatched rectangle (see Figure

1.1), where (2,1), (2,5),(3,5), (3,1).

Figure 1.1.

The following properties are easily verifiable:

a. ( ) ;

b. ( ) = ( ) ( ),

( ) = ( ) ( );

c. = ( = = ),

( ).

Algebraic problems and exercises for high school

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7. The intersection and the union of a family of sets. A family of

sets is a set {| } = {} whose elements are the sets ,

, (). We say that , , () is a family of sets

indexed with the set .

Let there be a family of sets {| }. Its union (or the union of

the sets , ) is the set

.

The intersection of the given family (or the intersection of the

sets , ) is the set

In the case of = {1,2, , }, we write

8. Euler-Wenn Diagrams. We call Euler Diagrams (in USA

Wenns diagrams) the figures that are used to interpret sets (circles,

squares, rectangles etc.) and visually illustrate some properties of

operations with sets. We will use the Euler circles.

Example. Using the Euler diagrams, prove Morgans law.

Solution.

In fig. 1.2.a, the hatched part is ; the uncrossed one

(except ) represents ( ).

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Figure 2.2.a Figure 3.2.b

In fig. 1.2b the side of the square hatched with \ \ \ \ is equal

to () and the one hatched with / / / / is equal to (). All the

hatched side forms () () (the uncrossed side is exactly

).

From these two figures it can be seen that ( ) (the

uncrossed side of the square in fig. 1.2.a coincides with ()

() (the hatched side of fig. 1.2.b), meaning

( ) = () ().

1.2. Solved exercises 1. For any two sets and , we have

= \ (\)

Solution. Using the definitions from the operations with sets, we

gradually obtain:

From this succession of equivalences it follows that:

which proves the required equality.

Algebraic problems and exercises for high school

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Remark. The equality can also be proven using Eulers diagrams.

So = \(\).

2. Whatever , are, the following equality takes place:

Solution. The analytical method. Using the definitions from

the operations with sets, we obtain:

This succession of equivalences proves that the equality from

the enunciation is true.

The graphical method. Using Eulers circles, we have

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Figure 4.3.a Figure 5.3.b

In fig. 1.3.a, we have ( ) ( ), which represents the

hatched side of the square. From fig. 1.3.b it can be seen that

( ) ( ) = ( ) ( ).

3. For every two sets , , the equivalence stands true:

Solution. Let \ = \. We assume that . Then there

is with or with .

In the first case we obtain \ and \ , which

contradicts the equality \ = \. In the second case, we obtain

the same contradiction.

So, if \ = \ = .

Reciprocally, obviously.

4. Sets = {1,2,3,4,5,6,7,8,9,10}, = {2,4,6,8,10} and =

{3,6,9} are given. Verify the following equalities:

a) \( ) = (\) (\);

b) \( ) = (\) (\).

Solution. a) We have = {2,3,4,6,8,9,10} , \( ) =

{1,5,7}, \ = {1,3,5,7,9}, \ = {1,2,4,5,7,8,10}, (\) (\) =

{1,5,7} = \( ).

Algebraic problems and exercises for high school

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b) For the second equality, we have

= {6} , \( ) = {1,2,3,4,5,6,7,8,9,10} , (\)

(\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = \( ).

5. Determine the sets A and B that simultaneously satisfy the

following the conditions:

1) = {1,2,3,4,5};

2) = {3,4,5};

3)1 \;

4) 2 \.

Solution. From 1) and 2) it follows that {3,4,5} and

{3,4,5} . From 3) it follows that 1 or 1 . If 1 ,

then from = {1,2,3,4,5} it follows that 1 . But, if 1 ,

then 1 , because, on the contrary it would follow that 1 =

{3,4,5}. So 1 and 1 remain. Similarly, from 4) it follows that

2 and so 2 . In other words, {3,4, 5} {2,3,4,5} and

{3, 4, 5} {1,3,4,5} with 2 , 1 and that is why

= {2,3,4,5}, and = {1,3,4,5}.

Answer: = {2,3,4,5}, = {1,3,4,5}.

6. The sets = {11 + 8 }, = {4 } and =

{11(4 + 1) 3 }, are given. Prove that = .

Solution. To obtain the required equality, we will prove the truth

of the equivalence:

.

Let . Then and and that is why two

integer numbers , , exist, so that = 11 + 8 = 4

11 = 4( 2). In this equality, the right member is divisible by 4,

and 11, 4 are prime. So, from 11 4 it follows that 4, giving =

4 for one . Then

= 11 + 8 = 11 4 + 8 = 11 4 + 11 3 = 11(4 + 1) 3

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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which implies , in other words, we have proven the implication

. (1)

Similarly, let . Then exists with

= 11(4 + 1) 3 = 11 4 + 11 3 = 11 4 + 8 = 4(11 + 2)

Taking 4 = and 11 + 2 = , we obtain

= 11 + 8 = 4 ,

which proves the truth of the implication

. (2)

From (1) and (2) the required equality follows.

7. The following sets are given

and = .

Prove that:

a) all sets with = {3,4,5,6,7,8,9};

b) all = { 2 }, so that = {3}.

Solution. We determine the set :

Then = (3; 6) = {3,4,5}.

a) All possible subsets of are

, {3}{4}{5}{3,4}{3,5}{4,5}{3,4,5} =

The required sets are such that = {3,4,5,6,7,8,9} and

will thus be like = {6,7,8,9}, where (), namely, the sets

required at point a) are:

1{6,7,8,9} , 2{3, 6, 7,8, 9} , 3{4, 6, 7,8, 9} , 4{5,6, 7,8,9} ,

5{3,4,6,7,8,9}, 6{3,5,6,7,8,9}, 7{4,5,6, 7,8,9}, 8{3,4,5,6, 7,8,9}..

b) Because , then and vice versa. Considering

that 2 = {3,4,5,6,7,8,9}, we obtain 2 {4,9}, namely 2

{3,2,2,3} = . The parts of the set are:

Algebraic problems and exercises for high school

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, {3} , {2} , {2} , {3} , {3,2} , {3,2} , {3,3} , {2,2} ,

{2,3}, {2,3}, {3,2,2}, {3,2,3}, {3,2,3}, {2,2,3}, .

From the condition = {3} it follows that is one of the

sets

1 = {3}, 2 = {3,3}, 3 = {2,3}, 4 = {2,3}, 5 = {3,2,3},

6 = {3,2,3}, 7 = {2,2,3}, 8 = = {3,2,2,3}.

Answer: a) {1, 2, 3, 4, 5, 6, 7, 8};

b) {1, 2, 3, 4, 5, 6, 7, 8}.

8. Determine ,, and , if

= {1,2,3,4,5,6} , = {1,2} , = {5,6} , =

= {3,4}.

Solution. From = = {3,4}, it follows that {3,4}

.

We know that

= ( ) ( ) = ( ) ( ),

= ( ) ( ) = ( ) ( ).

Then:

1 (1 1 ) ((1 1 )

1 ).

The following cases are possible:

a) 1 and1 ;

b) 1 and 1

(case no. 3, 1 and 1 1 = {3,4}, is impossible).

In the first case 1 and 1 , from = {5,6} it follows

that 1 , because otherwise 1 and 1 1

= {5,6}. So, in this case we have 1 = {3,4}which is

impossible, so it follows that 1 , 1 . Similarly, we obtain 2

and 2 and 2 , 5 and 5 ,5 and 6 , 6 , 6

.

In other words, we have obtained:

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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= {1,2,3,4}, = {3,4,5,6}, = {3,4} and = {1,2,5,6}.

Answer: = {1,2,3,4}, = {3,4,5,6}, = {3,4} and

= {1,2,5,6}.

9. The sets = {1,2}, = {2,3} are given. Determine the fol-

lowing sets:

a) ; b) ; c) 2;

d) 2; e) ( ) ( ); f) ( ) ;

g) ( ) ( ).

Solution. Using the definition for the Cartesian product with

two sets, we obtain:

a) = {(1,2), (1,3), (2,2), (2,3)};

b) = {(2,1), (2,2), (3,1), (3,2)};

c) 2 = {(1,1), (1,2), (2,1), (2,2)};

d) 2 = {(2,2), (2,3), (3,2), (3,3)};

e) ( ) ( ) = {(2,2)};

f) = {1,2,3}; ( ) = {(1,2), (1,3), (2,2)

(2,3), (3,2), (3,3)};

g) ( ) ( ) = {(1,2), (1,3), (2,2), (2,3),

(3,2), (3,3)} = ( ) .

10. The sets = {1,2, }, = {3,4, } are given. Determine

and , knowing that {1,3} {2,4} .

Solution. We form the sets and {1,3} {2,4} :

Because {1,3} {2,4} , we obtain

For = 3 and = 2, we have (3,2) .

Answer: = 3, = 2.

Algebraic problems and exercises for high school

23

11. If , then

Prove.

Solution. ( ) = and that is why

(we have used the equality ( ) = ( ) ( )).

12. How many elements does the following set have:

Solution. The set has as many elements as the fraction

(2 + 1) (22 + + 1) has different values, when takes the values

1,2,3, , 1000. We choose the values of for which the fraction takes

equal values.

Let , , < 1 with

Then

But and consequently

For = 2, we obtain = 3, and for = 3, we have = 2.

Considering that < , we obtain = 2 and = 3. So, only for =

2 and = 3, the same element of the set : = 5 11 is obtained.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Answer: the set has 999 elements, namely () = 999.

13. Determine the integer numbers , that make the following

statement true ( 1) ( 3) = 13.

Solution. We write

As 13 is a prime number, and , , the following cases are

possible:

meaning the proposition () is true only in these situations:

14. Determine the set

Solution. Because:

it follows that that equality

is possible, if and only if we simultaneously have:

Then:

a) if at least two of the numbers , and are different, we

cannot have the equality:

namely in this case = ;

b) if = = , then = and = {}.

Answer: 1) for = = , we have = {2}; 2) if at least two of

the numbers , and are different, then = .

Algebraic problems and exercises for high school

25

15. Determine the set

Solution. Possible situations:

+ 2 4 /3 or + 2 > 4 /3.

We examine each case:

In this case we have:

All integer numbers for which 1 3 2 applies, are:

1,0,1.

Then

All integer numbers for which 3 2 < x 9 applies, are:

2,3,4,5,6,7,8,9.

We thus obtain:

Answer: = {1,0,1,2,3,4,5,6,7,8,9}.

16. Determine the values of the real parameter for which the

set has:

a) one element;

b) two elements;

c) is null.

Solution. The set coincides with the set of the solutions of

the square equation

and the key to the problem is reduced to determining the values of the

parameter for which the equation has one solution, two

different solutions or no solution.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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a) The equation (1) has one (two equal) solution, if and only if

= 0 for = 1 0 or if = 1 = 0.

We examine these cases:

2) For = 1, the equation (1) becomes 5 + 15 = 0 = 3. So,

the set consists of one element for

b) The equation (1) has two different roots, if and only if > 0,

namely

c) The equation (1) doesnt have roots < 0 392

60 28 >

Answer:

17. Let the set = {3,4,5} and = {4,5,6}. Write the elements

of the sets 2 2 and ( ( )) ( ).

Solution. a) For the first set we have:

Algebraic problems and exercises for high school

27

2 = {(3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3), (5,4), (5,5)},

2 = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)},

2 2 = {(4,4), (4,5), (5,4), (5,5)}.

b) For the second set we have:

Then

Answer:

18. Given the sets = {1,2,3,4,5,6,7} and = {2,3,4}, solve

the equations = and ( ) = ().

Solution. In order to solve the enunciated equations, we use the

properties of the symmetrical difference: ( ) = , its

associativity and commutativity .

But

Consequently,

We calculate

Thus

Answer: = , = .

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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19. The following sets are given

Determine the sets , , , , , , ( )

( ) and ( ).

Solution. 1) We determine the sets and .

The inequation () is equivalent to the totality of the three

systems of inequations:

So

In other words,

2) We determine the required sets with the help of the graphical

representation

Algebraic problems and exercises for high school

29

20. 40 students have a mathematics test paper to write, that

contains one problem, one inequation and one identity. Three

students have successfully solved only the problem, 5 only the

inequation, 4 have proven only the identity, 7 have solved not only the

problem, 6 students not only the inequation, and 5 students have

proven not only the identity. The other students have solved

everything successfully. How many students of this type are there?

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Solution. Let be the set of students who have correctly solved

only the problem, only the inequation, that have proven only

the identity, the set of students that have solved only the problem

and the inequation, the set of students that have solved only the

problem and have proven the identity, the set of students who

have solved only the inequation and have proven the identity, and

the set of students that have successfully solved everything.

From the given conditions, it follows that () = 3, () = 5,

() = 4, () = 8, () = 7, () = 9.

But, as each of the students who have written the test paper

have solved at least one point of the test correctly and, because the

sets , , , , , , have only elements of the null set in common,

the union of the sets , , , , , , is the set of students that have

written the paper.

Consequently,

So

Answer: 4 students out of all that have taken the test have

solved everything successfully.

21. (Mathematician Dodjsons problem)

In a tense battle, 72 out of 100 pirates have lost one eye, 75

one ear, 80 one hand and 85 one leg. What is the minimum number

of pirates that have lost their eye, ear, hand and leg at the same time?

Solution. We note with the set of one-eyed pirates, with

the set of one eared pirates, with the set of pirates with one hand

and with the set of the pirates with one leg.

The problems requires to appreciate the set .

Algebraic problems and exercises for high school

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It is obvious that the universal set is composed from the set

and the number of pirates that have kept either both

eyes, or both ears, or both hands, or both legs.

Consequently,

It follows that the set is not smaller (doesnt have fewer

elements) than the sum of the elements of the sets , , , and

(equality would have been the case only if the sets , ,

and , two by two, do not intersect).

But,

Substituting, we have () = 100 namely 100 (

) + 30 + 25 + 20 + 15.

Consequently, ( ) 100 30 25 20 = 10.

Therefore, no less than 10 pirates have lost their eye, ear, hand

and leg at the same time

Answer: No less than 10 pirates.

22. Of a total of 100 students, 28 study English, 8 English and

German, 10 - English and French, 5 French and German, 3 students -

all three languages. How many study only one language? How many

not even one language?

Solution. Let be the set of students that study English, -

German, - French.

Then the set of students attending both English and German is

, English and French - , French and German - ,

English, German and French - , and the set of students

that study at least one language is .

From the conditions above, it follows that the students who only

study English form the set \( ) ( ) , only German -

\( ) ( ), only French - \( ) ( ).

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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But, because , we have (\(( ) ( ))) =

() (( ) ( )) = () (( ) + ( )

( )) = () ( ) ( ) + ( ).

Similarly, (\(( ) ( ))) = () ( )

( ) + ( ); (\(( ) ( )) = ()

( ) ( ) + ( ).

Let be the set of students who only study one language, then

Consequently,

= 28 + 30 + 42 2.8 2.10 2.5 + 3.3 = 63. () = 63.

The number of students who dont study any language is equal

to the difference between the total number of students and the

number of students that study at least one language, namely () =

() ( ), where is the set of students that dont study

any language and the set of all 100 students.

From problem 20 we have

So () = 100 80 = 20.

Answer: 63 students study only one language, 20 students dont

study any language.

Algebraic problems and exercises for high school

33

1.3. Suggested exercises 1. Which of the following statements are true and which are

false?

2. Which of the following statements are true and which are

false (, and are arbitrary sets)?

3. Let = { 2 12 + 35) = 0},

= { 2 + 2 + 35) = 0}; = { 2 + 2 + 35) = 0}

a) Determine the sets , and .

b) State if the numbers 1/5, 5, 7, 1/2 belong to these sets or

not.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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4. Determine the sets:

5. Let = { x = 4n + 6m,n,m };

a) Write three elements belonging to set .

b) Determine if 26, 28, 33 belong to .

6. Indicate the characteristic properties of the elements

belonging to the sets: = {4, 7, 10}, = {3, 6, 12} , =

{1, 4, 9, 16, 25}, = {1, 8, 27, 64, 125}.

7. How many elements do the following sets have?

8. Let there be the set = {3,2,1,1,2,3}. Determine the

subsets of :

9. Determine the sets:

Algebraic problems and exercises for high school

35

10. Compare (,,=,,) the sets and , if:

11. Let = {1, 2, 3, 4, 5, 6, 7}, = {5, 6, 7, 8, 9, 10}. Using the

symbols ,,, (complementary), express (with the help of , and

) the sets:

.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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12. Determine the set , in the case it is not indicated and its

parts , , simultaneously satisfy the conditions:

has more elements than ;

has more elements than ;

Algebraic problems and exercises for high school

37

13. Determine the sets , , , , , , ,

( ), , ( ), , , ( ) , ( ), if:

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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14. Let

a) Determine the sets:

b) How many elements does the set = {

(699 100 )} have?

15. Determine the sets , , if

16. Determine the set and its parts and , if

17. Compare the sets and , if

Algebraic problems and exercises for high school

39

18. Determine the values of the real parameter for which the

set has one element, two elements or it is void, if:

19. Determine the number of elements of the set :

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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20. The sets ,, are given. Determine .

21. Determine , if:

22. Prove the properties of the operations with sets.

23. Determine the equalities (, , etc. are arbitrary sets):

Algebraic problems and exercises for high school

41

24. Given the sets = {1, 2, 3}, = {2, 3, 4}, = {3, 4, 5} and

= {4, 5, 6}, write the elements of the following sets:

25. Given the sets , and , solve the equations () =

, where {,, }:

26. Determine the sets , , , , , ,

( ), ( ), if:

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Algebraic problems and exercises for high school

43

2. Relations, functions

2.1. Definitions and notations

Relations, types. Relations formation

Let and be two non-empty sets, and their Cartesian

product. Any subset is called a relation between the

elements of and the elements of . When = , a relation like

is called a binary relation on set .

If there exists a relation like , then for an ordered pair

(, ) we can have (, ) or (, ) . In the first case,

we write and we read " is in relation with ", and in the second

case , that reads " is not in relation with ". We hold that

(, ) .

By the definition domain of relation A X B, we understand

the subset consisting of only those elements of set that are in

relation with an element from , namely

By values domain of relation we understand the

subset consisting of only those elements of set that are in

relation with at least one element of , namely

The inverse relation. If we have a relation , then by

the inverse relation of the relation we understand the relation

1 as defined by the equivalence

namely

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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Example 1. Let us have = {1, 2}, = {4, 5, 6} and also the

relations

Determine the definition domain and the values domain of

these relations and their respective inverse relations.

Solution.

The formation of relations. Let , , be three sets and let us

consider the relations , . The relation

formed in accordance with the equivalence

namely

is called the formation of relation and .

Example 2. Let us have , , , , , from Example 1.

Determine

1 1 and ( )1.

Solution. We point out that the compound of the relations

with exists if and only if = .

Because , , it follows that and

doesnt exist.

We determine :

Algebraic problems and exercises for high school

45

(2,4) and {(4,1), (4,2)} {(2,1), (2,2)} ;

(2,6) , but in we dont have a pair with the first

component 6. It follows that

c) , exists. By repeating the

reasoning from b),

and 1 = {(2,2)}.

and 1 1 = {(1,2), (2,2)}.

The equality relation. Let be a set. The relation

is called an equality relation on . Namely,

Of a real use is:

1stTheorem. Let , , , be sets and , ,

relations. Then,

1) ( ) = ( ) (associativity of relations formation)

The equivalence Relations. The binary relation 2 is called:

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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a) reflexive, if whatever is;

b) symmetrical, if ( ), (), ;

c) transitive, if (( ) ), () , , ;

d) anti-symmetrical, if (( ) = ), () , , ;

e) equivalence relation on A, if it is reflexive, symmetrical and

transitive;

f) irreflexive, if whatever is.

Let be an equivalence relation on set . For each element

, the set

is called the equivalence class of modulo (or in relation to R), and

the set

is called a factor set (or quotient set) of through .

The properties of the equivalence classes. Let be an

equivalence relation on set and , . Then, the following

affirmations have effect:

Partitions on a set. Let be a non-empty set. A family of

subsets { } of is called a partition on (or of ), if the

following conditions are met:

2nd Theorem. For any equivalence relation on set , the factor

set / = { } is a partition of .

Algebraic problems and exercises for high school

47

3rd Theorem. For any partition = { } of , there exists

a sole equivalence relation on , hence

The relation is formed according to the rule

It is easily established that is an equivalence relation on and

the required equality.

Example 3. We define on set the binary relation according to

the equivalence ( ) , where , fixed, (),

.

a) Prove that is an equivalence relation on .

b) Determine the structure of the classes of equivalence.

c) Form the factor set /. Application: = 5.

Solution. a) Let , , . Then:

1) reflexivity = ;

2) symmetry

3) transitivity

From 1) - 3) it follows that is an equivalence relation on .

b) Let . Then

In accordance with the theorem of division with remainder for

integer numbers and , we obtain

Then

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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where = + , and therefore

where is the remainder of the division of through . But,

In other words, the equivalence class of coincides with the

class of the remainder of the division of through .

c) Because by the division to only the remainders 0, 1, 2, . . .,

1, can be obtained, from point b) it follows that we have the

exact different classes of equivalence:

The following notation is customarily used

. Then

where consists of only those integer numbers that divided by yield

the remainder

.

For n = 5, we obtain

with

Definition. The relation is called a congruency relation

modulo on , and class = is called a remainder class modulo

and its elements are called the representatives of the class.

The usual notation:

( ) (mode ) ( is congruent with

modulo ),and

Algebraic problems and exercises for high school

49

is the set of all remainder classes modulo .

Example 4 (geometric). Let be a plane and the set of all the

straight lines in the plane. We define as the binary relation in

accordance with

a) Show that is an equivalence relation on .

b) Describe the equivalence classes modulo .

c) Indicate the factor set.

Solution. a) It is obvious that is an equivalence relation (each

straight line is parallel to itself; if 1 2 2 1 and

b) Let . Then the class

consists only of those straight lines from that are parallel with the

straight line .

c) / = { } is an infinite set, because there are an

infinity of directions in plane .

Example 5. The set = {1,2,3,4,5,6, 7,8,9}is given and its parts

a) Show that = {1, 2, 3, 4 } is a partition of .

b) Determine the equivalence relation on .

c) Is = {1, 2, 3 } a partition on ?

Solution. a) 1) We notice that

meaning the system of the subsets of the set A defines a partition on

.

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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b) In accordance to the 3rd Theorem, we have

So:

= {( 1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (3,6), (4,3), (4,4), (4,6), (6,3),

(6,4), (6,6), (5,5), (5,7), (5,8), (8,5), (8,8), (8,7), (7,5), (7,7), (7,8), (9,9)}.

which proves that system does not define a partition on .

Order relations. A binary relation on the set is called an

order relation on , if it is reflexive, anti-symmetrical and transitive.

If is an order relation on , then 1 is also an order relation

on (verify!). Usually, the relation is denoted by "" and the

relation 1 by "", hence

With this notation, the conditions that "" is an order relation

on the set are written:

reflexivity ;

asymmetry ( ) = ;

transitivity ( ) .

Example 6. On the set we define the binary relation in

accordance with

Show that is an order relation on .

Solution. We verify the condition from the definition of order

relations.

1) Reflexivity

Algebraic problems and exercises for high school

51

2) Asymmetry. Let , with and . It follows that

there are the natural numbers , with = . and = . .

Then

which implies that

3) Transitivity. Let , , with and . Then there

exists , with = and = hence,

Because , , the implication is true

which proves that is an order relation on the set .

Remark. The order relation is called a divisibility relation on

and it is written as , namely

( reads "b divides a", and "a is divisible through b").

Functional relations Let and be two sets. A relation is called an

application or a functional relation, if the following conditions are met:

1) () , so as ;

An application (or function) is a triplet = (, , ), where

and are two sets and is a functional relation.

If is an application, then for each there exists,

according to the conditions 1) and 2) stated above, a single element

, so that ; we write this element with (). Thus,

The element () is called the image of element

through application , set is called the definition domain of noted

by () = , and set is called the codomain of and we usually say

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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that is an application defined on with values in . The functional

relation is also called the graphic of the application (function) ,

denoted, later on, by . To show that is an application defined on

with values in , we write : or , and, instead of

describing what the graphic of (or ) is, we indicate for each

its image (). Then,

i.e.

The equality of applications. Two applications = (, , )

and = (,, ) are called equal if and only if they have the same

domain = , the same codomain = and the same graphic

= . If , : , the equality = is equivalent to () =

(), () , that is to say

The identical application. Let be a set. The triplet (, , 1) is,

obviously, an application, denoted by the same symbol 1 (or ) and it

is called the identical application of set .

We have

Consequently,

1: and 1() = for () .

By (, ) we will note the set of functions defined on with

values in . For = , we will use the notation () instead of

(, ).

In the case of a finite set = {1, 2, , }, a function

: is sometimes given by means of a table as is:

Algebraic problems and exercises for high school

53

where we write in the first line the elements 1, 2, , of set , and

in the second line their corresponding images going through , namely

(1), (2), , ().

When = {1,2, , }, in order to determine the application,

we will use : and the notation

more frequently used to write the bijective applications of the set in

itself. For example, if = {1,2}, then the elements of () are:

If : , , , then we introduce the notations:

- the image of subset through the application .

Particularly, () = , the image of application ;

is the preimage of subset through the application .

Particularly, for , instead of writing 1({}), we simply

note 1(), that is

- the set of all preimages of through the application , and

- the complete preimage of set through the application .

The formation of applications. We consider the applications

= (, , ) and = (, , ), so as the codomain of to coincide

with the domain of . We form the triplet = (, , ).

Then is also an application, named the compound of the

application with the application , and the operation "" is called the

formation of applications. We have

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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namely

4th Theorem. Given the applications

. It follows that a) ( ) = ( )

(associativity of the formation of applications); b) 1 = 1 = .

Example 7. We consider the relations and

[0,+) [0,+), , defined as it follows: 2 = ,

= 2, = + 1.

A. Determine which of the relations , 1, , 1, , 1 are

functional relations.

B. Find the functions determined at point A.

C. Calculate , , , , 1, 1 (f, g, h are

the functions from point B.)

D. Calculate ( )( 3), (1 )(1/2), ( )(1/3).

Solution. We simultaneously solve and .

a) We examine the relation .

2 , i.e.

1) () = 2 , so that ;

which means that is a functional relation. We write the application

determined by through , = ( ,,).

b) We examine the relation 1 .

which means that if = { < 0}, then there is no ,

so that 1 , which proves that 1 is not a functional relation.

Algebraic problems and exercises for high school

55

c) For the relation , we have: and 1 are functional relations.

We denote by = ([0,+), [0,+), ),and 1 = ([0,+), [0,+),

1), the functions (applications) defined by and 1, respectively.

d) We examine the relation :

1) + 1 and so

so that ;

which proves that is a functional relation.

e) For the relation 1, we obtain:

1)() () = 1 , so that y1;

meaning that 1 is also a functional relation. We denote by =

(,, ) and 1 = (,, 1).

C. From A and B, it follows that:

a) Then , and dont exist, because the domains

and codomains do not coincide:

= (,, ), = ([0,+), [0,+), ), = (,,)).

b) We calculate , , 1 and 1 .

So and 1 = 1 = 1.

D. We calculate:

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Injective, surjective and bijective applications. An application

: is called:

1) injective, if

(equivalent: 1 2 (1) (2));

2) surjective, if

(any element from B has at least one preimage in );

3) bijective, if is injective and surjective.

Remark. To prove that a function : is a surjection, the

equation ( ) = must be solvable in for any .

Of a real use are:

5th Theorem. Given the applications

, we have:

a) if and are injective, then is injective;

b) if and are surjective, then is surjective;

c) if and are bijective, then is bijective;

d) if is injective, then is injective;

e) if is surjective, then is surjective.

6th Theorem. The application : with the graphic =

is a bijective application, if and only if the inverted relation 1 is a

functional relation (f1is an application).

This theorem follows immediately from

Inverted application. Let : be a bijective application

with the graphic = . From the 6th Theorem, it follows that the

triplet 1 = (, , 1) is an application (function). This function is

called the inverse of function .

We have

1: , and for ,

Algebraic problems and exercises for high school

57

i.e.

7th Theorem. Application : is bijective, if and only if

there exists an application : with = 1 and = 1.

In this case, we have = 1.

Real functions The function : is called a function of real variable, if

= () . The function of real variable : is called a real

function, if . In other words, the function : is called a

real function, if A and B . The graphic of the real

function : is the subset of 2 , formed by all the

pairs (, ) 2 with and = (), i.e.:

If the function is invertible, then

Traditionally, instead of 1() = we write = 1(). Then

the graphic of the inverted function 1 is symmetrical to the graphic

function in connection to the bisector = .

Example 8. For the function

the inverted function is

The graphic of function is the branch of the parabola = 2

comprised in the quadrant I, and the graphic of function 1 is the

branch of parabola = 2 comprised in the quadrant I (see below Fig.

2.1).

Algebra operations with real functions Let , : be two real functions defined on the same set.

We consider the functions:

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= + : ,

defined through () = () + (), () ;

= + g sum-function.

= : ,

defined through () = () (), () ;

= . product-function.

= : ,

defined through () = () (), () ;

= difference-function.

=

: 1,

defined through () =()

(), () 1

where 1 = { () };

=

quotient-function.

: ,

defined through () = () = {(), () 0,

(), () < 0,}

for () ;

- module-function.

Figure 2.1.

Algebraic problems and exercises for high school

59

Example 9. Let , : [0,+) with () = 2 and () =

. Determine , and /

Solution. As , share the same definition domain, the

functions , and / make sense and

Example 10. Let : [0,+), () = 2and : [0,+)

,() = . Determine a) and b) .

Solution. a) = makes sense and we obtain the function

: with

b) = : [0,+) [0,+), also makes sense, with

2.2. Solved exercises 1. Let = {2,4,6,8} and = {1,3,5,7}, , =

{(, )| 6 1} :

a) What is the graphic of the relation ?

b) Make the schema of the relation .

Solution.

a) Because , and , it follows that

So:

b) See Fig. 2.2 below.

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Figure 2.2.

2. Let = {1,2,3,4} and = {1,3,5,7,8}. Write the graphic of

the relation = {(, )3 + 4 = 10} .

Solution. We observe that 3 + 4 = 10 3 = 2(5

2) is even and (5 2) 3. Considering that , and ,

we obtain: (2,4), and (1,7). The equality 3 + 4 = 10 is

met only by = 2 and = 1. So = {(2,1)}.

3. Let = {1,3,4,5} and = {1,2,5,6}. Write the relation

using letters , and , if the graphic of the relation is known.

Solution. (, ) .

4. On the natural numbers set we consider the

relations , , , 2 , defined as it follows: (, ): =

{(3,5), (5,3), (3,3), (5,5)} ; ; = 12 ;

= 3.

a) Determine , , , , , , , .

b) What properties do each of the relations , , , have?

c) Determine the relations 1, 1, 1 and 1 .

d) Determine the relations , , 1 1, ( )1,

and 1 .

Algebraic problems and exercises for high school

61

Solution:

(for any natural numbers, there are natural numbers that exceed it).

(for any natural numbers there is at least one natural number that

exceeds it).

(for example, equation 2 = + 12 doesnt have solutions in ).

(for any , we have = 3/).

b) 1) is symmetrical, transitive, but is not reflexive. For

example (2,2) ; because (3,3) :, it follows that isnt

antireflexive either.

2) is - reflexive( , () ),

- anti-symmetrical (( )

( ) = ),

- transitive (( ) ).

So is an order relation on set .

3) is antireflexive ( , () ), anti-symmetrical

(( ) ( = 12 and = 12) =

= ), but it is not symmetrical ( = 12 =

12 ), and it is not transitive (( ) ( =

12 = 12) = 24 ).

4) is not antireflexive ( 3, () , 0 = 3.0), it is anti-

symmetrical (( ) ( = 3 = 3) = 9 =

0 = ), and for 0 , the equalities = 3 and = 3 cannot

take place), it is not reflexive (for example, 1 3.1 11), it is not

Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

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transitive (( ) ( = 3 = 3) = 9 3 ,

generally ).

So

i.e.

i.e.

i.e.

which implies

i.e.

In other words,

Algebraic problems and exercises for high school

63

+ 12 = 3, which implies

which implies

5. We consider the binary relation defined on as it follows:

a) Prove that is an equivalence relation.

b) Determine the factor set /.

Solution.

1) Because = , () , we have , () .

3) Let and , , , . Then

In other words, the binary relation is reflexive, symmetrical

and transitive, which proves that is an equivalence relation on .

b) Let . We determine the class of equivalence of in

connection to .

Then the set factor is

We observe that

In other words, for 1, we have 2 and therefore

= 2, and the factor can also be written

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6. Let = {1,2,3,4,5} and = {, , , , }. Which of the

diagrams in Fig. 2.3 represents a function defined on with values in

and which doesnt?

Solution. The diagrams a), c), e) reveal the functions defined on

with values in . Diagram b) does not represent a function defined

on with values in , because the image of 2 is not indicated.

Diagram d) also does not represent a function defined on with values

in , because 1 has two corresponding elements in , namely , .

7. Let A = {1, 2, 3} and = {, }. Write all the functions

defined on with values in , indicating the respective diagram.

Solution. There are eight functions defined on with values in .

Their diagrams are represented in Fig. 2.4.

Figure 2.3.

Algebraic problems and exercises for high school

65

Figure 2.4.

8. Let = {1, 2, 3, 4}, = {, , }, = {, , , } and =

{1, 2}.

a) Draw the corresponding diagrams for two surjective functions

defined on with values in.

b) Draw the diagram of the injective functions defined on with

values in .

c) Draw the diagram of a function denied on with values in ,

that is not injective.

d) Draw the corresponding diagrams for two bijective functions

defined on with values in .

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Solution. a) The diagrams of two surjective and two non-

surjective functions defined on with values in are represented in

Fig. 2.5 a) and b).

Figure 2.5.

b) The diagram of the injective functions defined on with

values in are represented in Fig. 2.6.

c) The diagram of a non-injective function defined on with

values in is represented in Fig. 2.7.

d) The diagrams of the two bijective functions defined on with

values in are represented in Fig. 2.8.

Algebraic problems and exercises for high school

67

Figure 2.5.

Figure 2.6.

Figure 2.7.

Figure 2.8.

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9. Let = {0, 1,5, 6}, = {0, 3, 8, 15} and the function

: as defined by the equality ( ) = 2 4 + 3, ()

.

a) Is the function surjective?

b) Is injective?

c) Is bijective?

Solution. a) We calculate () = {(0), (1), (5), (6)} =

{3,0, 8, 15} = . So, is surjective.

b) The function is also injective, because (0) = 3, (1) = 0,

(5) = 8 and (6) = 15, i.e. 1 2 (1) (2)

c) The function is bijective.

10. Using the graphic of the function : , : ,

show that the function is injective, surjective or bijective.

a) () = { 2, (, 2),

3 6, [2,+).

c) : [1,+] ,() = {

3, 1 < 0,, 0 < 1,

0.5( + 1), > 1.

Solution. If any parallel to the axis of the abscises cuts the

graphic of the function in one point, at most (namely, it cuts it in one

point or not at all), then the function is injective. If there is a parallel to

the abscises axis that intersects the graphic of the function in two or

more points, the function in not injective. If () is the values set of

the function and any parallel to the abscises axis, drawn through the

ordinates axis that are contained in (), cuts the graphic of the

function in at least one point, the function is surjective. It follows that

a function is bijective if any parallel to the abscises axis, drawn

through the points of (), intersects the graphic of in one point.

a) The graphic of the function is represented in Fig. 2.9.

Algebraic problems and exercises for high school

69

We have () = (,+) and any parallel = , to

the abscises axis intersects the graphic of the function in one point.

Consequently, is a bijectve function.

() = {2 6 + 8, 2 6 + 8 0,

2 + 6 8, 2 6 + 8 < 0=

= {2 6 + 8, if (,2] [4,+),

2 + 6 8, if (2,4).

Figure 2.9.

The graphic of the function is represented in Fig. 2.10.

We have () = [0; 3] [1; 3]. Any parallel = ,

(0,1) intersects the graphic of the function in four points; the

parallel = 1 intersects the graphic in three points; any parallel =

, (1; 3] intersects the graphic of the function in two points.

So the function is neither surjective, nor injective.

c) The graphic of the function is represented in Fig. 2.11. We

have

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Any parallel = to the abscises axis cuts the graphic of the

function () in one point, at most ( = 3, = 2, = 4) and that

is why () is injective. The equation () = has a solution

only for 3, hence () is not a surjective function.

d) We explain the function :

() = max( + 1, 1 ) = { + 1, 1 + 1,1 , + 1 < 1

=

= { + 1, 0,1 , < 0.

Figure 2.10.

Figure 2.11.

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71

The graphic of the function is represented in Fig. 2.12.

We have () = , () = [1,+). Any parallel = ,

(1,+) intersects the graphic in two points, therefore is not

injective. Straight line = 1/2 [0,+) doesnt intersect the

graphic of the function , so is not surjective either.

Figure 2.12.

11. Determine which of the following relations are applications,

and which ones are injective, surjective, bijective?

For the bijective function, indicate its inverse.

Solution.

Because

the equation 2 = + 3 doesnt have solutions in , it follows that

is not a functional relation.

Then, because () = = [1; 0], and

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with 3/2 3/2, and is not a functional relation.

c) We have: (, ) = () = 2. Then:

because 1, 2 (0,+);

4) The equation () = has solutions in [0,+), if and

only if 0, which proves that is not a surjection (number -2, for

example, doesnt have a preimage in [0,+).

d) We have:

Repeating the reasoning from c), we obtain that is an

injection. Furthermore, the equation () = [0,+) has the

solution = [0,+) , and that is why is a surjective

application. Then is a bijective application and, in accordance with

the 6th Theorem, the relation 1 is also an application (function).

In accordance with the same 6th Theorem, we have

12. Let { = 1,2,3,4,5,6,7,8,9} and () given; using the

following table:

determine:

a)

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73

b)

c) Calculate 1(1); 1(4); 1(7).

Solution.

13. Let = {1,2,3} and = {, , }. We examine the

relations

a) Determine , , , and , , .

b) Which of the relations , are applications? Indicate the

application type.

c) Determine the relations 1, 1 and 1. Which one is a

function?

d) Determine the relations 1, 1, 1, 1,

1 and 1. Which ones are applications?

Solution.

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b) is not an application, because element 3 has two images

and . is a bijective application; is an application, nor injective (the

elements 3 1 have the same image c), nor surjective (b doesnt have

a preimage).

We have:

1 is not an application, because the element has two

images 1 and 3; 1 is a bijective application; 1is not an application,

because 1 (the element doesnt have an image);

Only the relation 1is an application, neither injective, nor

surjective.

14. Given the functions: , : ,

() = {3 , 2, 1, > 2,

() = max( 2, 3 ),

show that = .

Solution. The functions and are defined on and have

values in . Lets show that () = (), () .

We explain the function :

() = {3 , 1 3 , 1, 3 < 1,

= {3 , 2, 1, > 2,

= ()

no matter what is. So = .

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75

15. The functions , : ,

() = { + 2, if 2,+10

3, if > 2,

() = { 2, if < 3,2 5, if 3

are given.

a) Show that and are bijective functions.

b) Represent graphically the functions and 1 in the same

coordinate register.

c) Determine the functions: = + , = , = . , = /.

d) Is function bijective?

Solution. a), b) The graphic of the function is represented in

Fig. 2.13 with a continuous line, and the graphic of 1 is represented

with a dotted line.

Figure 2.13.

We have:

1() = { 2, if 4,3 10, if > 4,

() = { + 2, if < 1,1

2( + 5), if 1.

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c) We draw the following table:

We have

() =

{

2, 2,4

3( + 1), 2 < < 3,

1

3(7 5), 3,

() =

{

4

2, 2,

2

3(8 ), 2 < < 3,

5

3(5 ), 3,

() =

{

2 4, 2,1

3(2 + 8 20), 2 < < 3,

1

3(22 + 15 50), 3,

() =

{

+ 2

2, 2,

+ 10

3( 2), 2 < < 3,

+ 10

3(2 5), 3.

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77

d) We have () = 4, () (,2], therefore is not

injective, in other words, is not bijective. This proves that the

difference (sum) of two bijective functions isnt necessarily a bijective

function.

16. Lets consider the function : {/} {/},

a) Show that function is irreversible.

b) Determine 1.

c) In what case do we have = 1?

Solution:

( )(1 2) = 0 1 = 2 is injective.

2) Let {/}. We examine the equation () = . Then

If

Imposibile. So ( )/( ) {/}, which proves

that the equation () = has solutions in {/} for any

{/}. This proves that is a surjective function.

From 1) - 2), it follows that is a bijective function, hence it is

also inversibile.

b) We determine

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c) In order to have = 1, it is necessary that the functions

and 1 have the same definition domain and so = . This

condition is also sufficient for the equality of the functions and 1.

It is true that, if = , the functions and 1are defined on

{/}, take values in {/}, and 1() =+

+= (), ()

{

}, namely = 1.

17. Represent the function () = 52 2 + 8 as a

compound of two functions.

Solution. We put

Then

Answer:

18. Calculate , , ( )(4) and ( )(4), where:

a) () =3

1 and () = ;

b) () = + 3 and () = 2 4;

c) () = {2 + 6, x 32 5, 3

and () = {5 2, 1,

2 2 + 4, > 1.

Indicate ( ) and ( ).

Solution: a) We have

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79

So ( )(4) ( )(4), which proves that the comutative

law of function composition, generally, does not take place:

.

c) To determine the functions and , in this particular

case we proceed as it follows:

( )() = ( )(()) == {2() + 6(), () < 3

2() 5, () 3=

= {

2() + 6() () < 3,

2() 5, () 3 and 1,

2(2 2 + 4) 5, () 3 and > 1

=

{

(5 2)2 + 6(5 2) 5 2 < 3,

2(5 2) 5, 5 2 3 and 1,

2(2 2 + 4) 5, 5 2 3 and > 1

=

= {

252 + 10 8 < 1/5,10 1, 1/5 1,

22 + 4 13, > 1.

( )() = (()) = {5(), () 1,

2() () + 4, () > 1.

We notice that

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So for () 1, we obtain

( )() = {5(2 + 6) 2, if [3 10,3)

5(2 5) 2, if [3,+).

Similarly,

and that is why for () > 1, we obtain

for (,3 10).

To sum it up, we have

( )() =

= {

4 + 123 + 342 12 + 4, (,3 10),

52 + 30 2, if [310,3),

10 7, if 3.

( )(4) = 10.4 27 = 67.

19. Solve the following equations:

a) ( )() = ( )() , if () = 3 + 1, () =

+ 3;

b) ( )() = , if () =+1

+, , {};

c) ( )() = ( )(), if () = 22 1, () = 43 3.

Solution. We determine the functions ( )() and (

)():

The equation becomes

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81

Answer: = 2.

b) We calculate ( )():

The equation becomes

Answer: {1,1}

c) We determine ( ) and ( )():

The equation becomes

meaning the equality is true for any .

Answer: .

20. Let , : given by () = 2 + + 12 and () =

2 + 2. Show there is no function : , so that

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Solution. The relation () can also be written

We suppose there is a function : that satisfies the

relation (). We put in () = 1 and = 1. We obtain:

which implies (4) +

(2) (2) + (4), contradiction with the hypothesis. So there is

no function with the property from the enunciation.

21. Determine all the values of the parameters , for

which ( ) = ( )() , () , where () = 2 and

() = 2 + + .

Solution. We determine and :

Then

Answer: = 1, = 0, () = 2 = ().

22. Given the functions , , ,

.

Show that:

a) is not injective;

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83

b) is injective;

c) is bijective and determine 1.

Solution. a) Let (1) = (2).

Then

For example, () = 3( + 4) + 3, taking 1 = 0 and 2 = 4,

we obtain (0) = (4) = 3, 1 2.

because

so is an injective function.

We prove that is surjective. Let . We solve the equation

(the root of an uneven order exists out of any real number). So is a

surjection, therefore, is a bijective function.

We determine 1:

23. Let : [1,+) [1,+) with

a) Prove that is a bijective function.

b) Determine 1.

Solution. a) We have () = (2 + 1) 3. We represent this

function as a compound of two functions:

, : [1,+) [1,+), where () = 3, () = 2 + 1.

Then () = ( )() , which implies that = is bijective,

being the compound of two bijective functions.

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b) 1 = ( )1 = 1 1. We determine 1 and 1.

Because:

We solve this equation according to [1,+).

The discriminant is:

The equation has only one root in [1,+):

Then

So,

= (1 + 4 33

)/2, with 1: [1,+) [1,+).

24. Considering the function : with the properties:

a) Determine the function .

b) Calculate (1998).

Solution. a) For 2 = 0 from 1), it follows that (1) = (1) +

(0), which implies that (0) = 0. For 2 = 1 from 1), we obtain

Then

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85

Let {0, 1}. Then

On the other hand, 1

1=1+

1= 1 +

1 implies

From (2) and (3) it follows that

So () = , () .

Answer: a) () = ; b) (1998) = 1998.

25. Using the properties of the characteristic function, prove the

equality:

Solution. Using the properties = = , we will

prove the required equality by calculating with the help of , and

the characteristic functions of sets ( ) and ( )

( ):

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which implies ( ) and ( ) ( ).

2.3. Suggested exercises 1. Determine the definition domains and the values domains of

the following relations:

2. Let = {2,4,6,8} and = {1,3,5,7}, .

a) Determine the graphic of the relation .

b) Construct the schema of the relation :

1) = {(, ) < 3 and > 3};

2) = {(, ) > 2 and < 5};

3) = {(, ) > 6 or > 7};

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87

3. Let = {1,2,3,4} and = {1,3,5,7,8}. Write the graphic of

the relation , if:

4. Let = {, 3, 4, 5}, = {, 2, 5, 6} and be the graphic of

the relation . Write the relation with propositions containing letters

and , with and :

5. Let A = {, 2.3,4}. Analyze the properties of the relation

2(var. 1-6) and 2 (var. 7-14):

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7) = {(, ) 2 > 1 and > 1 };

8) = {(, ) 2{ > 0, > 0

or { < 0, < 0

};

9) = {(, ) 2 0 or 0 };

10) = {(, ) 2 1 or 1 };

6. For each binary relation defined on set :

a) determine the definition domain and the values domain ;

b) establish the properties (reflexivity, ireflexivity, symmetry,

anti-symmetry, transitivity);

c) determine the inverse relation 1(, ):

7. Given the set and the binary relation 2 , prove that is

an equivalence relation and determine the factor set /.

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89

8. Given the set = {1, 2,3,4,5,6,7,8, 9} and the subset system

= { , = 1, }, prove that defines a partition on and build

the equivalence relation .

9. We define on the binary relation :

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a) Show that is an equivalence relation on .

b) Determine the equivalence classes.

10. We define on the binary relation :

.

a) Show that is an equivalence relation.

b) Determine the equivalence classes.

11. Let 2 , where = {5,4,3,2,1, 0, 1, 2, 3, 4, 5, 6,

7, 8} with

a) Is an equivalence relation?

b) If so, determine the equivalence classes.

12. Determine: , , 1, , 1, 1 , if:

13. Determine the relations

, , 1, 1, 1 , ( )1 ;

5) = {(, ) 2 s even},

= {(, ) 2 is uneven};

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91

14. Let = {1, 2, 3, 4}, = {, , , } . Determine the

definition domain and the values domain of each of the following

relations , . Which ones are applications? Determine the application

type. Determine the relation 1 , 1, 1, 1. Are

they applications?

15. We consider the application : , () = sin .

Determine:

16. The application : is given by the table below, where

= {1, 2, 3, 4, 5, 6, 7, 8, 9} and = {, , , , , }.

Determine:

17. We consider the application : , ( ) = [] ([] is

the integer part of ). Determine

1({2,4,5}) and 1(1).

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18. The following application is given: : , () = 2.

Determine () and 1(), if = {1, 2, 3, 4, 5, 6 , 7, 8, 9, 10}.

19. Let and be two finite sets, || = . ||. How many

surjective applications : are there, if:

20. Given the graphic of the relation , establish if is a func-

tion. Determine and .

Changing and , make become an application injective,

surjective and bijective.

Draw the graphic for the relation 1. Is relation 1 a function?

What type is it?

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93

21. Let = {1, 2, 3, 4}, = {0, 1, 5, 6} and = {7, 8, 9}.

a) Draw the diagrams of two injective and two non-injective

functions defined on with values in .

b) Draw the diagrams of two surjective functions and two non-

surjective applications defined on with values in .

c) Draw the diagrams of two bijective and two non-bijective

functions defined on with values in .

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22. Let = {1, 3, 5, 6} and = {0, 1, 2, 3, 5}, , .

Which of the relations stated below represents a function defined on

with values in ? And one defined on with values in ? For functions,

indicate their type:

23. Using the graphic of the function : , show that is

injective, surjective or bijective. If bijective, determine 1and draw

the graphic of and 1in the same register of coordinates..

24. Which of the following relations 2 are functions?

Indicate the definition domain of the functions. Establish their type:

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95

25. Given the relation with = [3; 5] and = [4; 7]:

a) Does pair (4, 5) belong to the relation ? Why?

b) Indicate all the ordered pairs (, ) with = . Explain.

26. Given the function (), calculate its values in the indicated

points.

4) () = {2 5, > 0,2 + 3, 0,

(2), (0), (5);

5) () = {2 + 1, > 0 4, = 01 2, < 0

(5), (1), (1/2);

27. Determine (), if:

28. Given the functions () and (), determine the functions

+ , , and / , indicating their definition domain.

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29. Represent the function () as a compound of some

functions:

30. Given the functions () and ( ), determine the

functions and and calculate ( )(3) and ( )(3) in

options 1) 6) and ( )( 1) and ( )( -1) in options 7)

12).

31. Given the functions () = 2, () = 3 and () =

1, calculate:

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97

32. Determine if in the function pairs and one is the inverse

of the other:

33. The value of function is given. Calculate the value of the

function 1, if:

34. Determine 1, if:

35. The function: : is given:

a) Prove that is bijective.

b) Determine 1

c) Calculate 1 1 .

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36. Given the functions ( ) and ( ), determine the functions

( )() and ( )()(, : ):

37. Prove the equality of the functions and :

38. Determine the functions = + , = , = .

and = /:

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99

39. Let = {1, 2, 3, 4}, = {0, 1, 3, 4} and the functions

Can the functions , be defined? If so, determine

these functions. Make their diagrams.

40. a) Show that the function is bijective.

b) Determine 1.

c) Graphically represent the functions and 1 in the same

coordinate register.

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41. Show that the function : , () = 2 6 + 2

admits irreversible restrictions on:

Determine the inverses of these functions and graphically

represent them in the same coordinate register.

42. Using the properties of the characteristic function, prove the

following equalities (affirmations):

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101

3. Elements of combinatorics

3.1. Permutations. Arrangements. Combinations. Newtons Binomial

In solving many practical problems (and beyond), it is necessary

to:

1) be able to assess the number of different combinations

compound with the elements of a set or a multitude of sets;

2) choose (select) from a set of objects the subsets of elements

that possess certain qualities;

3) arrange the elements of a set or a multitude of sets in a

particular order etc.

The mathematical domain that deals with these kind of

problems and with the corresponding solving methods is called

combinatorics. In other words, combinatorics studies certain

operations with finite sets. These operations lead to notions of

permutations, arrangements and combinations.

Let = {1, 2, , } be a finite set that has elements. Set

is called ordered, if each of its elements associates with a certain

number from 1 to , named the element rank, so that different

elements of associate with different numbers.

Definition 1. All ordered sets that can be formed with elements

of given set (( ) = ) are called permutations of elements.

The number of all permutations of elements is denoted by the

symbol and it is calculated according to the formula:

By definition, it is considered 0 = 0! = 1 = 1! = 1.

Definition 2. All ordered subsets that contain elements of set

with elements are called arrangements of elements taken at

a time.

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The number of all arrangements of elements taken at a

time is denoted by the symbol and it is calculated with the formula

Definition 3. All subsets that contain elements of set with n

elements are called combinations of elements taken at a time.

The number of all combinations of elements taken at a time

is denoted by the symbol and it is calculated with the formula

where , ; 0 .

Remark. In all the subsets mentioned in definitions 1 - 3, each

element of the initial set appears only once.

Along with the combinations in which each of the different

elements of a set participates only once, we can also consider

combinations with repetitions, that is, combinations in which the same

element can participate more than once.

Let groups of elements be given. Each group contains certain

elements of the same kind.

Definition 1. Permutations of elements each one containing

1 elements 1, 2 elements 2 , elements , where 1 + 2 +

+ are called permutations of elements with repetitions.

The number of all permutations with repetitions is denoted by

the symbol 1,2,, and it is calculated using the formula:

Definition 2 . The arrangements of elements, each one

containing elements, and each one being capable of repeating itself

in each arrangement for an arbitrary number of times, but not more

than times, are called arrangements of elements taken at a

time with repetitions.

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103

The number of all arrangements with repetitions of elements

taken at a time is denoted by the symbol and it is calculated with

the formula

Definition 3'. The combinations of elements each one

containing elements, and each one being capable of repeating itself

more than once, but not more than times, are called combinations

of elements taken at a time with repetitions.

The number of all combinations with repetitions is denoted by

the symbol and it is calculated with the formula

In the process of solving combinatorics problems, it is

important to firstly establish the type (the form) of combination. One

of the rules used to establish the type of combination could be the

following table

It is often useful to use the following two rules:

Caution is advised to the order the elements are arranged in:

Arrangements if not all of

the elements from the

initial set participate;

Permutations if all

elements of the initial

set participate;

If the order isnt important,

we have combinations;

If the order is important,

we have arrangements

or permutations:

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The sum rule. If the object can be chosen in ways and the

object in n ways, then the choice "either A or B" can be made in +

ways.

The multiplication rule. If the object can be chosen in

ways and after each such choice the object can be chosen in ways,

then the choice "A and B" in this order can be made in . ways.

We will also mention some properties of combinations, namely:

The formula

is called Newtons Binomial formula ( ).

The coefficients 0,

1, 2, ,

from Newtons Binomial

formula are called binomial coefficients; they possess the following

qualities:

V. The binomial coefficients from development (7), equally set

apart from the extreme terms of the development, are equal among

themselves.

VI a. The sum of all binomial coefficients equals 2.

VI b. The sum of the binomial coefficients that are on even

places equals the sum of the binomial coefficients that are on uneven

places.

VII. If is an even number (i.e. = 2), then the binomial

coefficient of the middle term in the development (namely ) is the

biggest. If is uneven (i.e. = 2 + 1 ), then the binomial

Algebraic problems and exercises for high school

105

coefficients of the two middle terms are equals (namely =

+1)

and are the biggest.

VIII. Term , namely the ( + 1) term in equality (7),

is called term of rank + 1 (general term of development) and it is

denoted by +1. So,

IX. The coefficient of the term of rank + 1 in the

development of Newtons binomial is equal to the product of the

coefficient term of rank multiplied with the exponent of in this

term and divided by , namely

3.2. Solved problems 1. In how many ways can four books be arranged on a shelf?

Solution. As order is important and because all the elements of

the given set participate, we are dealing with permutations.

So

Answer: 24.

2. A passenger train has ten wagons. In how many ways can the

wagons be arranged to form the train?

Solution. As in the first problem, we have permutations of 10

elements of a set that has 10 elements. Then the number of ways the

train can be arranged is

Answer: 3 628 800.

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3. In how many ways can 7 students be placed in 7 desks so that

all the desks are occupied?

Solution.

Answer: 5 040.

4. How many phone numbers of six digits can be dialed:

1) the digit participates in the phone number only once;

2) the digit participates more than once?

(The telephone number can also start with 0.)

Solution. We have 10 digits on the whole: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

As the phone number can also start with 0, we have:

1) arrangements of 10 digits taken 6 at a time, i.e.

2) because the digit can repeat in the number, we have

arrangements with repetitions, i.e.

Answer: 1) 151 200; 2) 1 000 000.

5. The volleyball team is made out of 6 athletes. How many

volleyball teams can a coach make having 10 athletes at his disposal?

Solution. As the coach is only interested in the teams

composition it is sufficient to determine the number of combinations

of 10 elements taken 6 at a time, i.e.

Answer: 210.

6. How many 5 digit numbers can be formed with digits 0 and 1?

Solution. As the digits are repeating and their order is important,

we consequently have arrangements with repetitions. Here, = 5,

= 2.

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107

From digits 0 and 1 we can form 25 numbers of five digits.

But we have to take into consideration that the number cannot

begin with the digit zero. So, from the number 25 we have to begin

with zero. Numbers of this type are 24 . Therefore, the number we

seek is

Answer: 16.

7. How many three digit numbers can be formed with the digits

1, 2, 3, 4, 5, if