Tutorial

Permutations:

The ways in which a number of given objects can be arranged by taking all of

them or a specified number of objects out of them are called

PERMUTATIONS.   Thus the number of permutations of three objects, viz.  a,

b, and c, taking all of them at a time is 6 i.e., abc, acb, bcd, bac, cab and

cba.

The number of ways in which 2 objects can be taken and arranged out of 3

objects a, b and c is 6, viz.  ab, ba, bc, cb, ac and ca.   The number of

permutations of r things our of n things is denoted by npr.

Formulae:

1. n! = n(n-1)(n-2)....(1)

2. npr = n!/(n-r)!

Example: 10P4 = 10 x 9 x 8 x 7 = 5040

3. The number of ways in which n objects can be arranged in a circle is (n-1)!

Combinations:

The ways in which a specified number of objects can be taken out of a given

number of objects (without regard to their arrangements) are called

Combinations.  The symbol nCr denotes the number of combinations or r

things out of n things.  Thus, for example the number of combinations of 2

objects out of three given objects a, b and c is 3, viz.  ab, ca, bc.

Formulae:

1. nCr = nPr/r!

2. nCr = nCn-r

3. nCr + nCr-1 = (n+1)Cr

4. nC0 + nC1 + nC2 + nC3 + ......nCn = 2n

Circular Permutations

There are two types of circular permutations.

(a) If clockwise and anti clock-wise orders are different, then total number of

circular-permutations is given by (n-1)!

(b) If clock-wise and anti-clock-wise orders are taken as not different, then

total number of circular-permutations is given by (n-1)!/2!

Number of circular-permutations of n different things taken r at a time:-

(a) If clock-wise and anti-clockwise orders are taken as different, then total

number of circular-permutations = nPr /r

(b) If clock-wise and anti-clockwise orders are taken as not different, then

total number of circular permutation = nPr/2r

Examples

Example 1: In how many ways can the letters of the word ABACUS be

rearranged such that the vowels always appear together?

Solution: ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants

and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a"

appears twice.

Hence, the total number of rearrangements in which the vowels appear

together are (4! x 3!)/2!

Example 2: How many different four letter words can be formed (the words

need not be meaningful) using the letters of the word "MEDITERRANEAN"

such that the first letter is E and the last letter is R?

Solution: The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining

5 letters.

The second and third positions can either have two different letters or have

both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different

letters from the 8 available different choices. This can be done in 8 * 7 = 56

ways.

Case 2: When the two letters are same. There are 3 options - the three can

be either Ns or Es or As. Therefore, 3 ways.

Total number of posssibilities = 56 + 3 = 59.

Example 3: How many different signals can be made by 5 flags from 8-flags

of different colours?

Solution:  Number of ways taking 5 flags out of 8-flags  = 8P5

=   8!/(8-5)!      

=  8 x 7 x 6 x 5 x 4 = 6720

Example 4: A child has 3 pocket and 4 coins. In how many ways can he put

the coins in his pocket?

Solution: First coin can be put in 3 ways, similarly second, third and forth

coins also can be put in 3 ways.

So total number of ways = 3 x 3 x 3 x 3   = 34   = 81

Assignment

Discuss and post solution

1. A man has nine friends, four boys and five girls. In how many ways can he

invite them, if there have to be exactly three girls in the invitees?

2. A company manufactures pencils in boxes of 6, 9, and 20. The boxes are

sealed and the pencils cannot  be  sold  loose.  What is  the  largest  number 

of  pencils  that  a  wholesaler  cannot  purchase using some combination of

these boxes?

3. The number of ways in which 10 candidates A1, A2, ……, A10 can be

ranked so that A1 is always above A2 is ?

4. A hosted a party and invited all her friends and asked them to invite their

friends.There are n people in the party.Only S is not known to A.Each pair

that does not include A or S has exactly 2 common friends.Also,S knows

everyone except A.If only 2 friends can dance at a time,how many dance

numbers will be there at the party?

5. In a classroom there are 14 students seated in 3 rows of 5 chairs. The

place at the centre of the room is unoccupied. A teacher decides to reassign

the seats such that each student will occupy a chair  adjacent to his/her

present one (i.e. move one desk forward, backward, right or left). In how

many ways can this reassignment be done?

6. Consider a 4 digit number. the first 2 digits are equal and last 2 digits are

also equal. How many of such digits are perfect square?

7. Suppose u have a currency, named x, in 3 denominations, 1, 10 and 50. In

how many ways can 107 x be given in this currency?

Last Modified: Tuesday Aug 25, 2009 - 01:47. Revision: 9. Release Date: Friday Jul 3, 2009 - 22:00.

Probability

Table of ContentsTutorial Examples Assignment

Tutorial

An experiment is an act for which the outcome is uncertain. Examples of

experiments are rolling a die, tossing a coin, surveying a group of people on

their favorite soft drink, etc...

An experiment is said to be a random experiment, if it's out-come can't be

predicted with certainty.

Example; If a coin is tossed, we can't say, whether head or tail will appear.

So it is a random experiment.

A sample space S for an experiment is the set of all possible outcomes of

the experiment such that each outcome corresponds to exactly one element

in S.  The elements of S are called sample points.  If there is a finite number

of sample points, that number is denoted n(S), and S is said to be a finite

sample space.

For example, if our experiment is rolling a single die, the sample space would

be S = {1, 2, 3, 4, 5, 6}. If our experiment is tossing a single coin, our

sample space would be S = {Heads, Tails}. 

Every subset of a sample space is an event. It is denoted by 'E'. e.g. In

throwing a dice S={1,2,3,4,5,6}, the appearance of an event number will be

the event E={2,4,6}.

An event, consisting of a single sample point is called a simple event. e.g.

In throwing a dice, S={1,2,3,4,5,6}, so each of {1},{2},{3},{4},{5} and

{6} are simple events.

Compound event: A subset of the sample space, which has more than on

element is called a mixed event.e.g. In throwing a dice, the event of

appearing of odd numbers is a compound event, because E={1,3,5} which

has '3' elements.

Equally likely events: Events are said to be equally likely, if we have no

reason to believe that one is more likely to occur than the other. e.g. When a

dice is thrown, all the six faces {1,2,3,4,5,6} are equally likely to come up.

Exhaustive events: When every possible out come of an experiment is

considered. e.g. A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of

events.

Probability of an Event

 

If 'S' be the sample space, then the probability of occurrence of an event 'E'

is defined as:

P(E) = n(E)/N(S) = (number of elements in 'E'/ (number of elements in

sample space 'S')

Empirical Probability

Finding the probability of an empirical event is specifically based on direct

observations or experiences.

For example, a survey may have been taken by a group of people.  If the

data collected is used to find the probability of an event tied to the survey, it

would be an empirical probability.   Or if a scientist did research on a topic

and recorded the outcome and the data from this is used to find the

probability of an event tied to the research, it would also be an empirical

probability.

Equiprobable space

A sample space S is called an equiprobable space if and only if all the

simple events are equally likely to occur. e.g. A toss of a fair coin.  It is

equally likely for a head to show up as it is for a tail. 

Mutually Exclusive

Events E and F are said to be mutually exclusive if and only if   they have no

elements in common.

E.g. if the sample space is rolling a die, where S = {1, 2, 3, 4, 5, 6}, and E is

the event of rolling an even number, E = {2, 4, 6} and F is the event of

rolling an odd number, F = {1, 3, 5}, E and F are mutually exclusive,

because they have NO elements in common. 

Bayes' Theorem The short form of Bayes' Theorem states that if E and F are events, then

P(F|E) = P(E|F)P(F)

----------------------

P(E|F)P(F) + P(E|F')P(F')

Properties of Probability

1. 0 <= P(E) <= 1

2. P(not E) = 1 - P(E) So if, P(E) = 1/4 then P(not E) = 3/4.

3. "Or" probabilities with mutually exclusive events P (A or B) = P(A) +

P(B)

4. "Or" probabilities with events that are NOT mutually exclusive P (A or B)

= P(A) + P(B) - P(A AND B)

5. A and B are Independent Events if an only if P(A AND B) = P(A)P(B)

Examples

Example 1: Find the probability of getting a tail in tossing of a coin.

Solution: Sample space S = {H,T}  and n(s) = 2

              Event 'E' = {T}  and n(E) = 1

              therefore P(E) = n(E)/n(S) = 1/2

Example 2: A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles.

If a single marble is chosen at random from the jar, what is the probability of

choosing a red marble? a green marble? a blue marble? a yellow marble?

Solution: Outcomes:         The possible outcomes of this experiment are red,

green, blue and yellow.

Probabilities:      

P(red)      =      number of ways to choose red/total number of marbles =

6/22 = 3/11

 

P(green)      =      number of ways to choose green/total number of marbles

= 5/22

 

P(blue)      =      number of ways to choose blue/total number of marbles =

8/22 = 4/11

 

P(yellow)      =      number of ways to choose yellow/total number of marbles

= 3/22

Example 3: A man can hit a target once in 4 shots. If he fires 4 shots in

succession, what is the probability that he will hit his target?

Solution: The man will hit the target even if he hits it once or twice or thrice

or all four times in the four shots that he takes.

So, the only case where the man will not hit the target is when he fails to hit

the target even in one of the four shots that he takes.

The probability that he will not hit the target in one shot = 1 - 1/4 = 3/4

Therefore, the probability that he will not hit the target in all the four shots

=3/4 x  3/4 x 3/4 x 3/4 = 81/256

Hence, the probability that he will hit the target at least in one of the four

shots = 1 - 81/256

= 175/256 .

Example 4: What is the probability that the position in which the consonants

appear remain unchanged when the letters of the word Math are re-

arranged?

Solution: The total number of ways in which the word Math can be re-

arranged = 4! = 4*3*2*1 = 24 ways.

Now, if the positions in which the consonants appear do not change, the first,

third and the fourth positions are reserved for consonants and the vowel A

remains at the second position.

The consonants M, T and H can be re-arranged in the first, third and fourth

positions in 3! = 6 ways without the positions in which the positions in which

the consonants appear changing.

Therefore, the required probability = 3!/4! = 6/24 = 1/4

Assignment

Discuss and post solutions

1. Out of two-thirds of the total number of basket-ball matches, a team has

won 17 matches and lost 3 of them. What is the maximum number of

matches that the team can lose and still win three-fourths of the total

number of matches, if it is true that no match can end in a tie?

2. 4 people played a game of chess, where each one plays every other

player. What is the maximum number of points that any player could gather

if every win gets him 1 point ?

3. From a pack of 52 cards, all face cards are removed and four cards are

drawn. Then the probability that they are of different suit and different

denomination is

4. Five balls of different colours are to be placed in three boxes of different

sizes. Each box can hold all five balls. The number of ways in which we can

place the balls in the boxes so that no box remains empty is ?

5. How many arrangements can be made of the letters of the word

DRAUGHT the vowels never being separated?

6. If the integers m and n are chosen at random from integers from integers 

1 to 100 with replacement, then the probability that a no. of the form 7m + 7n

is divisible by 5 equals?

Functions

Table of ContentsTutorial Assignment

Tutorial

A relation is a set of ordered pairs where the first components of the

ordered pairs are the input values and the second components are the

output values.

A function is a relation that assigns to each input number EXACTLY ONE

output number.

The domain is the set of all input values to which the rule applies.  These

are called your independent variables. These are the values that

correspond to the first components of the ordered pairs it is associated with.

The range is the set of all output values.  These are called your dependent

variables. These are the values that correspond to the second components

of the ordered pairs it is associated with.

Function Notation

f(x) read "f of x"

f is the function name.  Output values are also called functional values.

Note that you can use any letter to represent a function name, f is a very

common one used.

x is your input variable. 

Think of functional notation as a fancy assignment statement.  

When you need to evaluate the function for a given value of x, you simply

replace x with that given value and simplify.  For example, if we are looking

for f(0), we would plug in 0 as the value of x in our function f.

If the function is constant, that means that the functional value

never changes, it is always equal to that constant.

f(x) = c, where c is a constant.

Assignment

Discuss and post solution

DIRECTIONS for Questions 1 and 2: Answer the questions based on the

following information:-

A, S, M and D are functions of x and y, and they are defined as follows:

    A(x, y) = x + y

    S(x, y) = x - y

    M(x, y) = xy

    D(x, y) = x/y, where y in not equal 0.

1. What is the value of M(M(A(M(x, y), S(y,x)), x), A(y, x)) for x = 2, y = 3 ?

2. What is the value of S(M(D(A(a, b), 2), D(A(a, b),2)), M(D(S(a, b), 2), D(S(a,

b),2))) ?

3. What is the value of 1.1!+2.2!+3.3!+--------+n.n! ?

4. Let g(x) be a function such that g(x + 1) + g(x − 1) = g(x) for every real x.

Then for what value of p is the relation g(x + p) = g(x) necessarily true for

every real x?

5. A function y = f(n) is defined, for all natural numbers, as the sum of the

digits of n.

if k is a natural number such that f(f(f(f(k)))) = 1 and k > f(f(k)).f (f(f(k))) > 1

what is the least number of digits that k can have?

6. A function y = f(n) is defined, for all natural numbers, as the sum of the

digits of n. for a natural number m, what is  the value of f(f(m − f(m)))?

Inequalities

Table of ContentsTutorial Examples Assignment

Tutorial

The term inequality is applied to any statement involving one of the

symbols <, >, , .

Example of inequalities are:

i.       x 1

ii.      x + y + 2z > 16

iii.     p2 + q2 1/2

iv.     a2 + ab > 1

Properties of Inequalities

1.    If a b and c is any real number, then a + c b + c.

e.g. -3 -1 implies -3+4 -1 + 4.

2.    If a b and c is positive, then ac bc.

e.g. 2 3 implies 2(4) 3(4).

3.    If a b and c is negative, then ac bc.

e.g. 3 9 implies 3(-2) 9(-2).

4.    If a b and b c, then a c.

e.g. -1/2 2 and 2 8/3 imply -1/2 8/3.

 

Solution of Inequality

By solution of the one variable inequality 2x + 3 7 we mean any

number which substituted for x yields a true statement.

e.g. 1 is a solution of 2x + 3 7 since 2(1) + 3 = 5 and 5 is less than and

equal to 7.

By a solution of the two variable inequality x - y 5 we mean any

ordered pair of numbers which when substituted for x and y, respectively,

yields a true statement.

e.g. (2, 1) is a solution of x - y 5 because 2-1 = 1 and 1 5.

Examples

Example 1: If -1 < x < 4 the determine a, b in a < 2x + 3 < b.

Solution:  We have -1 < x < 4 so multiply everything by 2. We will get

               -2 < 2x < 8

              Now add 3 to everything. so 1 < 2x + 3< 11.

Therefore a = 1, and b = 11.

Assignment

Discuss and post solution

1. If |b| > 1 and x = |a|b, then which one of the following is necessarily true?

a) a − xb <= 0 b) a − xb > 0 c) a − xb < 0 d) a − xb >= 0

2. If a, b, c are real numbers such that a < b < c and a + b + c = 6, ab + bc

+ ca = 9, then which among the following is definitely true?

(a) 0 < a < 1 (b) 1 < b < 3 (c) 3 < c < 4 (d) All of them (e) none of them

3. Let x be a real number such that 1 − 1/n < x <= 3 + 1/n is true for all

natural numbers n. Which among the following best describes x?

(a) 1 < x < 3 (b) 1 <= x <= 3 (c) 0 < x < 4 (d) 1 < x <= 3 (e) 0 < x <= 4

4. If a2b3c = 256/27 find min value of a + b + c, given a, b, c are positive real

nos.

5. From four positive real nos a, b, c and d, 4 distinct combination of sum of

three numbers are formed S1, S2, S3 and S4. IF abcd = 5. FInd the min value

of S1S2S3S4.

Binomial Theorem

An algebraic expression consisting of two terms with a positive or negative

sign between them is called a binomial expression. e.g.  (a+b), ( P / x2)  –

(Q / x4) etc.

Binomial Theorem: When a binomial expression is raised to a power ‘n’ we

would like to be able to expand it. The binomial theorem assists us in doing

this. It converts such an expression into a series.

Binomial Theorem for positive integral index:

(x+y)n = xn + nC1xn-1y+nC2xn-2y2+-----+nCrxn-ryr+ -------+---------+nCn-1xyn-1   + ncnyn.

(x+y)n = nr=0∑nCrxn-ryr

therefore Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :

(1+x)n = nCoxo+nC1x+nC2x2+---------+nCrxr+------+nCn-1xn-1+nCnxn.

Properties of Binomial – Expansion (x+y)n :

(i)        There are (n+1) terms in the expansion.

(ii)        In each term, sum of the indices of ‘x’ and ‘y’ is equal to ‘n’.

(iii)        In any term, the lower suffix of ‘c’ is equal to the index of ‘y’, and the

index of x = n-(lower suffix of c).

(iv)             Because  nCr = nCn-r,

    so we have :

nCo = nCn

nC1=nCn-1

nC2=nCn-2 etc.

It follows that the coefficients of terms equidistant from the beginning and

the ends are equal.

General Terms : (r +1) th term from beginning in

      (x+y)n is called general – term, and

it is denoted by

                        Tr+1 = nCrxn-ryr

Greatest Coefficient : In any binomial expansion middle-term has the

greatest.

Coefficient. So

(i)       If ‘n’ is even, then greatest – coefficient  =  nCn/2

(ii)       If ‘n’ is odd, then greatest – coefficients are nC(n+1)/ 2  and  nC(n-1)/2

Properties of Binomial coefficients :

(1) The sum of binomial coefficient in (1 + x)n  is 2n.

(2)  The sum of the coefficients of the odd-terms in (1+x)n is equal to the

sum of coefficients of the even terms and each is equal to 2n-1.

Progressions

Table of ContentsArithmetic Progression Geometric Progression Harmonic Progressions Video Tutorials

 

Arithmetic Progression

An arithmetic progression is a sequence of numbers such that the difference

of any two successive members of the sequence is a constant.

For example, the sequence 3, 5, 7, 9, 11,... is an arithmetic progression with

common difference 2.

Arithmetic progression property:

a1 + an = a2 + an-1 = ... = ak+an-k+1

Formulae for the n-th term can be defined as:

an = 1/2 x (an-1 + an+1)

If the initial term of an arithmetic progression is a1 and the common

difference of successive members is d, then the n-th term of the sequence is

given by

an = a1 + (n - 1)d, n = 1, 2, ...

The sum S of the first n values of a finite sequence is given by the formula:

S = 1/2(a1 + an)n, where a1 is the first term and an the last.

or

S = 1/2(2a1 + d(n-1))n

 

Example 1: Find the sum of the first 10 numbers from this arithmetic

progression 1, 11, 21, 31...

Solution: we can use this formula S = 1/2(2a1 + d(n-1))n

              S = 1/2(2.1 + 10(10-1))10 = 5(2 + 90) = 5.92 = 460

Example 2: The sum of the three numbers in A.P is 21 and the product of

their extremes is 45. Find the numbers.

Solution: Let the numbers are be a - d, a, a + d

Then a - d + a + a + d = 21

3a = 21

a = 7

and (a - d)(a + d) = 45

a2 - d2 = 45

d2 = 4

d = +2

Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2.

In both the cases numbers are the same.

Geometric Progression

A geometric progression is a sequence of numbers such that the quotient

of any two successive members of the sequence is a constant called the

common ratio of the sequence.

a geometric sequence can be written as:

aq0=a, aq1=aq, aq2, q3, ... where q ≠ 0, q is the common ratio and a is a scale

factor.

Formulae for the n-th term can be defined as:

an = an-1.q

an = a1.qn-1

The common ratio then is:

q = ak

ak-1

A sequence with a common ratio of 2 and a scale factor of 1 is 1, 2, 4, 8, 16,

32...

A sequence with a common ratio of -1 and a scale factor of 3 is 5, -5, 5, -5, 5,

-5,...

If the common ratio is:

Negative, the results will alternate between positive and negative.

Greater than 1, there will be exponential growth towards infinity (positive).

Less than -1, there will be exponential growth towards infinity (positive

and negative).

Between 1 and -1, there will be exponential decay towards zero.

Zero, the results will remain at zero

Geometric Progression Properties

a2k = ak-1.ak+1

a1.an = a2.an-1 =...= ak.an-k+1

Formula for the sum of the first n numbers of geometric progression

Sn = a1 - anq 1  -  q  = a1. 1 - qn 1 - q

Infinite geometric series where |q| < 1

 

If |q| < 1 then an -> 0, when n -> ∞ So the sum S of such a infinite geometric

progression is:

S = 1

1 - x

which is valid only for |x| < 1

Harmonic Progressions

A sequence of numbers is said to form a harmonic progression if their reciprocals

form an arithmetic progression.

e.g. 1/4, 1/9, 1/14, 1/19 are in H.P. since 4,9, 14, 19 are in A.P.

In general, the numbers 1/a, 1/(a+d), 1/(a+2d), ..., 1/(a+(n+1)d) are in H.P.

Note:

i) The series formed by the reciprocals of the terms of a geometric series is

also a geometric series.

ii) There is no general method of finding the sum of a harmonic progression.