**The Advanced Guide to ****GMAT**** Math **

**This supplementary Guide is an absolute MUST if you aiming at 99 Percentile Math
**

**The Advanced 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.**

*Interested in GMAT Algebra, Arithmetic, Set Theory, Geometry & Coordinate Geometry?** Click here for Essential Guide to GMAT Math *

*Interested in Math Question Bank with over 400 Tough Math Problems covering All areas of GMAT Math? **Click here for Math Question Bank for GMAT Winners*

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 Advanced 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*

*Check out our Math Special Offer & get a 30% Discount click here for Details.*

**Got Questions? Check out our FAQ Page here**

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**Try this eBook for a FULL TWO MONTHS at OUR COST**

**If you are not happy with this guide for whatever reason, you can ask for a full refund, NO QUESTIONS ASKED at anytime.**

**Just Quote the Paypal Receipt Number and the money will be refunded to you. No questions asked!! **

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__An Excerpt from the Advanced 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

2.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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