standard prep books.

- 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.

personal GMAT prep strategy

that got me a

how it can benefit YOU.

An Excerpt from the 'Winners' Guide to GMAT Math- Part IIFactorialThe 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 PermutationsSuppose 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. CombinationsIf 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 Rules1. 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 Examplei) 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) SolutionThe 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) SolutionThe 11 white balls can be arranged in 11! waysThe 9 black balls can be arranged in the 12 places in 12! ways. Total number of arrangements = 11!12!/3! ExampleIn 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? SolutionEach 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. ExampleHow many 4 digit numbers can be formed from the, digits 1, 5, 2, 4, 2, 9, 0, 4, 2i) 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 Grammar Problems? Click here for an exclusive ebook for students who want an in-depth understanding of how the GMAT Grammar works, with over 50 unique Sentence corrections questions with explanations.Click here for the Winners' Guide to GMAT RC: Winning techniques to triple your reading speed, and application of these techniques to crack the GMAT RC. |

I got an Quant percentile of just 70 in 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: 95Percentile) |

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! not only a good sum of money, but also got the highest score possible! - Ruchi K (GMAT Quant: 99 Percentile) |

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700+ GMAT Prep

700+ GMAT Prep