The Winners' Guide to GMAT Math - Part II
This supplementary Guide is an absolute MUST if you aiming at 700+
The Winners' Guide exclusively covers four GMAT Math topics that are not covered well in the
standard prep books.
These are: Number Theory, Statistics, Probability, Permutation/Combination
PLUS an exclusive list of Over One Hundreds formulae covering ALL sections of the GMAT.
You cannot afford to miss these if you want to be a Math Winner.
In recent years, probability & Combinations questions have been appearing with increasing
regularity in the GMAT, more so at the 700+ levels.
Most recent test takers have faced 2-4 probability/combination questions in their GMAT Exam.
This is one subject that is unfortunately, not covered well in the commonly available GMAT
prep books.
And some of the specialist books go way above the level required for the GMAT, resulting in
wasted effort that does not result in a higher score.
Here is some study material on these Topics that is GMAT specific and should help you with
your GMAT prep.
If you want to do further practice on these topics, you should definitely go through the 'Winners
Guide to GMAT Math'. This guide exclusively covers these four topics in detail.
Unique Features of the Winner's Guide to GMAT 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 GMAT 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 GMAT. No learning
difficult concepts or theories that will never get tested on the GMAT.
- 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 GMAT problems are based
around these.
If you are already familiar with the usage of these formulae, you should be able to tackle any
GMAT Quant problem with confidence and ease.
Most GMAT 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 90 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 GMAT 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 GMAT
questions are based around.
Most GMAT 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-gmat.
Now it is being made available for anyone who is aiming at 700+ score in the GMAT.
ALL this for a special Introductory Price of Only USD 25.00
(After completing your transaction, you will be taken to a page where you can instantaneously download the eBook)
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Click here to read how I got a 780 score in the GMAT.
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I got an Quant percentile of just 70 in
my first attempt. The main reason
was that my fifth question was on
probability theory. What took the cake
was that I got another two
probabilty/perm question within the
first 20.
I finally decided to leave nothing to
chance and prepared these tough
topics well with the help of 'Winners
guide to GMAT Math'.
Entered the exam room with full
confidence and came out with a 95
percentile in math and an overall
score of 720.
Jim Pankhurst (GMAT Quant: 95
Percentile)
My friends had warned me about the
dreaded
probability/permutation/combn
questions.
I spent a lot of time looking around
for a good book on these topics, that
targeted the GMAT.
My search finally ended with the
Winners' Guide and that helped me
get an amazing 97 percentile in the
Quant section.
A 'Must Read' if you want to make
sure you do your best in the Quant
section.
Jo Birks (GMAT Quant: 97 Percentile)
Aiming at a 99 percentile 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,
but also got the highest score
possible!
Ruchi K (GMAT Quant: 99 Percentile)

An Excerpt from the 'Winners' Guide to GMAT 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 =
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
\ objects can be partitioned in r x r x r .... n times = rn ways
Example
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
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
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.
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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