Winners' Guide to  99 Percentile New GRE Math
This Comprehensive Guide is the  Final Word as far as your GRE Math prep is concerned.
An absolute MUST if you aiming at a perfect score in GRE Math.

The Winners' Guide exclusively covers four GRE Math topics that are not covered well in the standard
prep books.
These are:
Number Theory, Statistics, Probability, Permutation/Combination
Aiming at a perfect GRE Math
score, I wanted to cover ALL areas
well. One of the areas that required
extra work was Perm/Comb and
Stats.
I almost paid a private tutor $500 to
teach me these topics, when I
stumbled upon the 'Winners' Guide.
Spent sometime with it, and quickly
realized that I did not really need
any private coaching at all!

Thanks to this guide, I managed to
save not only a good sum of money,
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Judy K (GRE Math 800)
I am an Engineer and was
expecting a perfect Math score.
However, I was shocked to see a
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GRE!!

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available guides that I had used are
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S. Suresh (GRE Math 800)
While I had a decent understanding
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I stumbled upon this site while
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Interested in GRE Algebra, Arithmetic, Set Theory, Geometry & Coordinate Geometry?
Click here for Winners' Guide to CORE New GRE Math

Interested in Math Question Bank with over 500 Tough Math Problems covering All areas of GRE Math?
Click here for Math Question Bank for New GRE Winners
In recent years, and in the New GRE, probability & Combinations questions have been appearing
with increasing frequency.

Most recent test takers have  faced  2-4 probability/combination questions in their GRE Exam.

This is one subject that is unfortunately, not covered well in the commonly available GRE prep books.

And some of the specialist books go way above the level required for the GRE, resulting in wasted
effort that does not result in a higher score.

Here is some study material  on these Topics that is GRE specific and will help you with your GRE
prep.

If you want to do further practice on these topics, you should definitely go through the 'Winners Guide
to 99 Percentile GRE Math'. This guide exclusively covers these four topics in detail.


Unique Features of the  Winner's Guide to 99 Percentile GRE Math

- Focus on the difficult GMAT Math topics that are not covered well in Standard books.
(Four Topics are covered:
Number Theory, Statistics, Probability, Permutation/Combination)

- All Theory & Questions based around the
Actual GRE questions that have appeared on
these topics in the recent past.

-
130 pages with Over One hundred Solved problems with detailed explanations.

-
No Superfluous Material. You study ONLY what is required for the GRE. No learning
difficult concepts or theories that will never get tested on the GRE.

-
Instant Delivery: Since this is an eBook, you will be able to download it instantaneously after
you have made the payment.

- A Comprehensive list of over One Hundred Formulae covering the following Topics:

I. Algebraic Formulae
II. Even and Odd Numbers
III. HCF & LCM
IV. Surds and Indices
V. Percentage
VII. Simple Interest and Compound Interest
VIII. Quadratic Equations
IX. Averages
X. Time, Speed and Distance
XI. Progression
XII. Series
XIII. Permutations and Combinations
XIV. Co-ordinate Geometry
XV. Probability
XVI. Set Theory
XVII Plane Figures
XVIII Solid Figure
XIX Conversions

Of course you do not have to memorize all these formulae by heart. It is enough if you can
derive them from some of the basic formulae.

Familiarity with these formulae is
Critical because a majority of GRE problems are based
around  these.

If you are already familiar with the usage of these formulae, you should be able to tackle any
GRE Quant  problem with confidence and ease.

Most GRE Winners  know at Least 90% of these formulae by heart and are able to derive
the others.
This provides a tremendous
sense of confidence going into the exam.



Why Should you Buy this Guide?

If you are aiming at 99 Percentile  in the GMAT Math, you need to cover ALL the Topics well.

There have been innumerable instances, where students get a Probability or a Stats Question
in the first 10.

As you well know, getting a single question wrong in the first 10 can jeopardise your chances
of a good score.

This guide has been specifically written to supplement the Standard prep Books available in
the market.

It includes questions similar to the ones that have appeared in the GRE in the recent past,
especially at the 700+ level.

You will NOT find these questions either in the Official Guide or in the Standard Prep Books.

In fact, these topics are pretty much simply ignored because they are usually (but Not always)
asked at a higher score level, and the official guide and the standard prep books target the
600-650 score audience.

Also, the Formulae in the guide comprise a comprehensive list of formulae that the GRE
questions are based around.

Most GRE Winners  know  these formulae by heart and are well versed with how to apply
them.
This gives them a Big
advantage during the exam.

While the commonly available books provide a list of formulae, they are by no means
comprehensive, more so since these books are
Not targeted at a 700+ audience.

This  guide was originally prepared exclusively for students who take private coaching from
my-prep.

Now it is being made available for anyone who is aiming at a 99 percentile score in the GRE.

ALL this for a special Introductory Price of Only USD 24.95

















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An Excerpt from the Winners' Guide to  99 Percentile New GRE Math

Factorial

The factorial of a number is the product of all the positive integers from 1 upto the number. The
factorial of a given integer n is usually written as n! and n! denotes the product of the first n
natural number. -
n! = n x (n – 1) x (n – 2) x ……… x 1
n! = n (n – 1)
0! = 1 as a rule.
Note :        Factorial is not defined for improper fractions or negative integers.



Permutation

If r objects are to be chosen from n, where n ≥ 1 and these r objects are to be arranged, and the
order of arrangement is important, then such an arrangement is called a permutation of n
objects taken r at, a time.
Permutations is denoted by nPr or (n, r)


e.g., If it is required to seat 5 men and 4 women in a row such that women occupy the even
places, in how many ways can this be done?

In a row of 9 positions, there are four places, and exactly 4 women to occupy them, which is
possible in 4! ways. The remaining S places can be filled up by 5 men in 5! ways,
Total number of seating arrangements = 4! 5! = 24 x 120 = 2880


Important Permutation Rules:

(i) The total number of arrangements of n things taken r at a time in which a particular thing
always occurs.

e.g., The number of ways in which 3 paintings can be arranged in an exhibition from a set of
five, such that one is always included.
number of ways        3. 5-1P3-1 = 3.4P2 = 36       
or        3! (4C2) = 6.6 = 36

(ii) The total number of permutations of n distinct things taken r at a time in which a particular
thing never occurs = n-1Pr

e.g., The number of ways in which 3 paintings from a set of five, can be displayed for a photo-
shoot, such that one painting is never picked.
= 5-1P3 = 4P3 ways = 24
It can be observed that
rn-1Pr-1 + n-1Pr = nPr

(iii) The number of permutations of n different objects taken r at a time, when repetitions are
allowed, is nr. The f place can be filled by any one of the n objects in ‘n’ ways. Since repetition is
allowed the second place can be filled in ‘n’ ways again. Thus, there are n x n x n r times ways =
nr ways to fill first r positions.



Circular Permutations

Suppose four numbers 1, 2, 3, 4 are to be arranged in the form of a circle.

The arrangement is read in anticlockwise direction, starting from any point as 1432, 4321, 3214
or 2143.
These four usual permutation correspond to one circular permutation.
Thus circular permutations are different only when the relative order of objects to be arranged is
changed.

Each circular permutation of n objects corresponds to n Linear permutations depending on
where (of the n positions) we start.

This can also be though of as keeping the position of one out of n objects fixed and arranging
remaining n – 1 in (n – 1)! ways.




Combinations

If r objects are to be chosen from n, where r ≤ n and the order of selecting the r objects is not
important then such a selection is called a combination of n objects taken r at a time and
denoted by  


In a permutation the ordering of objects is important while in a combination it is immaterial. e.g.,
AB and BA are 2 different Permutations but are the same combination.
Usually (except in trivial cases) the number of permutations exceeds the number of
combinations. Trivial cases are when r = 0 or 1.

e.g., If there are 10 persons in a party, and if every two of them shake hands with each other,
how many handshakes happen in the party?

SoIn: When two persons shake hands it is counted as 1 handshake and not two hence here we
have to consider only combinations.
2 people can be selected from 10 in 10C2 ways.
Hence, number of handshake =  10C2


Combinatorial Identities:

1. nCr = nCn – r
2. nCo = nCn = 1
3. n+1Cr = nCr + nCr – 1
4. n+1Cr+1 = nCr+1 + n–1Cr + n–1Cr-1
5. nPr = r! nC
6. The total no. of combinations of ‘n’ things taken some or all at a time nc = nC1 + nC2 + ……
nCn = 2n – 1


Important Combination Rules

1. The number of combinations of ‘n’ things taken ‘r’ at a time in which p particular thin will
always occur = n-pCr-p P things are definitely selected in 1 way. The remaining r – p things can
be selected from n – p things in n-pCr-p ways.

In how many ways can 7 letters be selected from the alphabet such that the vowels are always
selected.
Soln : There are 5 vowels a, e, i, o, u which are selected in 1 way then possible number of ways
= 26-5C7-5 = 21C2

The number of combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things never
occur is n-pCr (n – p ≥ r)
p things are never to be selected.
Hence r things are to be selected from n - p in n–pCr ways It is clear that     n - p ≥ r for this to be
possible.
e.g. In how many ways can 7 letters be selected from the alphabet such that the vowels are
never selected.
Soln : As vowels (a, e, i, o, u) are never selected. The 7 letters can be selected from (20 – 5),
letters in = 26–5C7 = 21C7


3. The number of ways of dividing (partitioning) •n distinct things into r distinct groups, such that
some groups can remain empty = rn
One object ran be put into r partitions in r ways
n objects can be partitioned  r x r x r .... n times = rn ways



Examples
i)        In how many ways can 11 identical white balls and 9 black bells be arranged in a row so
that no two black balk are together?
Solution
The 11 white balls can be arranged in 1 way (all are identical)
The 9 black balls can be arranged in the 12 places in 12P9   ways

ii) In how many ways can they be arranged if black balls were identical? (all other conditions
remaining same)

Solution i
The 11 white balls can be arranged in 1 way.
The 9 black balls can be arranged in the 12 places in 12C9 ways.
Thus number of arrangements = 12C9
iii) In how many ways can they be arranged if all the balls are different. (all other conditions
remaining same)

Solution ii
The 11 white balls can be arranged in 11! ways
The 9 black balls can be arranged in the 12 places in 12! ways.
Total number of arrangements =  11!12!/3!   


Example
In a multiple choice test there are 50 questions each having 4 options, which are equally likely.
In how many ways can a student attempt the questions in the test?
Solution
Each question can be attempted in 4 ways and not attempted in 1 way. \Each question can be
attempted or unattempted in 5 ways.
Thus 50 questions can be attempted or attempted in 550 ways.
This will include the case when no questions are attempted.
\The student can attempt the paper in (5 to the power of 50) – 1  ways.

Example
How many 4 digit numbers can be formed from the, digits 1, 5, 2, 4, 2, 9, 0, 4, 2
i)        with repetition of digits.
ii)        without repetition of digits.


Solution
i) In the given set 4 is repeated twice and 2 thrice
\Number of distinct digits = 6
The 4 digit number can be formed in 5.63 ways when repetition is allowed.
Position I can be filled in 5 ways, (as it cannot have O)
The remaining 3 positions can be filled in 6 ways each.
Hence number of numbers = 5.63 = 1080
ii).   Position I can be filled in 5 ways.
Position II can be filled in 5 ways (it can contain any of 5 digits except the one in position 1
Thus number of such numbers = 5 x 5 x 4 x 3 = 300
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