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Permutations and Combinations Basics

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Posted over 3 years ago by Suresh

FUNDAMENTAL PRINCIPLE OF COUNTING

If an operation can be performed in 'm' different ways and another operation in 'n'different ways then these two operations can be performed one after the other in'mn' waysIf an operation can be performed in 'm' different ways and anotheroperation in 'n' different ways then either ofthese two operations can be performedin 'm+n' ways.(provided only one has to be done)

This principle can be extended to any number of operations

FACTORIAL 'n'

The continuous product of the first 'n' natural numbers is called factorial n and is deonoted by n! i.e, n!=1Ã—2Ã—3xâ€¦..x(n-1)xn.

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PERMUTATION

An arrangementthat can be formed by taking some or all of a finite set of things (orobjects) is called a Permutation.Order of the things is very important in case of

permutation.A permutation is said to be a Linear Permutation if the objects arearranged in a line. A linear permutation is simply called as a permutation.A permutationis said to be a Circular Permutation if the objects are arranged in the form of acircle.The number of (linear) permutations that can be formed by taking r things at a time

from a set of n distinct things is denoted by .%

NUMBER OF PERMUTATIONS UNDER CERTAIN CONDITIONS

1. Number of permutations of n different things, taken r at a time, when a particular thng

is to be always included in each arrangement , is .

2. Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each

arrangement is .

3. Number of permutations of n different things, taken all at a time, when m specified things always come together is

.

4. Number of permutations of n different things, taken all at a time, when m specified never come together is

.

5. The number of permutations of n dissimilar things taken r at a time when k(< r) particular things always occur is

.

6. The number of permutations of n dissimilar things taken r at a time when k particular things never occur is .

7. The number of permutations of n dissimilar things taken r at a time when repetition of things is allowed any number oftimes is

8. The number of permutations of n different things, taken not more than r at a time, when each thing may occur any

number of times is .

9. The number of permutations of n different things taken not more than r at a time .

*PERMUTATIONS OF SIMILAR THINGS*The number of permutations of nthings taken all tat a time when p of them are all alike and the rest are all different is

.If p things are alike of one type, q things are alike of other type, r things are alike

of another type, then the number of permutations with p+q+r things is .

CIRCULAR PERMUTATIONS

}1. The number of circular permutations of n dissimilar things taken r at a time is .

2. The number of circular permutations of n dissimilar things taken all at a time is .

3. The number of circular permutations of n things taken r at a time in one direction is .

4. The number of circular permutations of n dissimilar things in clock-wise direction = Number of permutations in

anticlock-wise direction = .

COMBINATION

A selection that can be formed by taking some or all of a finite set of things( or objects) is called a Combination

The number of combinations of n dissimilar things taken r at a time is denoted by .

1.

2.

3.

4.

5. The number of combinations of n things taken r at a time in which

a)s particular things will always occur is .

b)s particular things will never occur is .

c)s particular things always occurs and p particular things never occur is .

DISTRIBUTION OF THINGS INTO GROUPS

1.Number of ways in which (m+n) items can be divided into two unequal groups containing m and n items is

.

2.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and

the order of the groups is not important is

3.The number of ways in which mn different items can be divided equally into m groups, each containing n objects and

the order of the groups is important is .

4.The number of ways in which (m+n+p) things can be divided into three different groups of m,n, an p things

respectively is

5.The required number of ways of dividing 3n things into three groups of n each = .When the order of groups

has importance then the required number of ways=

DIVISION OF IDENTICAL OBJECTS INTO GROUPS

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more

items is

}The number of non-negative integral solutions of the equation .

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is

The number of positive integral solutions of the equation .

The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the

coefficient of in the expansion

he number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on, such that one

object of each kind may be included is the coefficient of is the coefficient of in the expansion

.

%{font-family:verdana}+*TOTAL NUMBER OF COMBINATIONS*+%

%{font-family:verdana}1.The total number of combinations of

things taken any number at a time when things are alike of

one kind, things are alike of second kindâ€¦. things are alike of kind, is

.%

%{font-family:verdana}2.The total number of combinations of

things taken one or more at a time when things are alike ofone kind, things are alike of second kindâ€¦. things are alike of kind, is%

.

SUM OF THE NUMBERS

Sum of the numbers formed by taking all the given n digits (excluding 0) is

Sum of the numbers formed by taking all the given n digits (including 0) is

Sum of all the r-digit numbers formed by taking the given n digits(excluding 0) is

%

%{font-family:verdana}Sum of all the r-digit numbers formed by taking the given n digits(including 0) is

DE-ARRANGEMENT:

The number of ways in which exactly r letters can be placed in wrongly addressed envelopes when n letters are placed

in n addressed envelopes is .

The number of ways in which n different letters can be placed in their n addressed envelopes so that al the letters are in

the wrong envelopes is .

IMPORTANT RESULTS TO REMEBER

In a plane if there are n points of which no three are collinear, then

1. The number of straight lines that can be formed by joining them is .

2. The number of triangles that can be formed by joining them is .

3. The number of polygons with k sides that can be formed by joining them is .

In a plane if there are n points out of which m points are collinear, then

1. The number of straight lines that can be formed by joining them is .

2. The number of triangles that can be formed by joining them is .

3. The number of polygons with k sides that can be formed by joining them is .

Number of rectangles of any size in a square of n x n is

In a rectangle of p x q (p < q) number of rectangles of any size is

In a rectangle of p x q (p < q) number of squares of any size is

n straight lines are drawn in the plane such that no two lines are parallel and no three lines three lines are concurrent.

Then the number of parts into which these lines divide the plane is equal to .

Image Credits: cristic, cosmolallie, farouqtaj, churl

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sanjucuckoo – over 2 years ago

concise andgood collection . one example of each type would make it complete.

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LinkSuresh – over 2 years ago

dear sanjucuckoo, we will have one more lesson P&C with solved examples, where all these concepts will be

revised….Keep watching


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LinkRuchi – over 2 years ago

These are basics!! Whats advanced then :(he he he…. j/k

nice compilation of stuff, am waiting for the solved examples..


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pradeepdemi – over 2 years ago

hi its very good i need probabilty with examples


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Linkravi_shashank – over 2 years ago

hi your explanation is excellent.and can you please explain even probability like this


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vsagashe – over 2 years ago

Excellent..!! Probably some examples would make it more useful for all viewers..


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swayam – over 2 years ago

This is tremendously good yarvery nice n easy explaination


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Aditi Mishra – over 2 years ago

this lesson helped me to clear my doubts but it will be more helpful with examples


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hamsini – over 2 years ago

where do i get exercises to solve on my own??


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Suresh – over 2 years ago

Dear hamsini, please go through our tests that are available in different test prep communities like CAT Prep,

GMAT Prep. In fact in the very near future we are going to come up with tests exclusively for Campus

Placements


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santosh gupta – over 2 years ago

u r gr8 sir


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Sudharshan Annamalai – over 2 years ago

Good work !! It would have been better with an example for each concept.

Thanks !!


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pujii – over 2 years ago

hi it is excellent i want more details about the usage of permutations and combinations in our daily life.


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Prashanth Kumar R – over 2 years ago

Very good lesson. Thank you :)


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rajpatti – over 2 years ago

excellent explanation on each topic great job thank u


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abinash – over 2 years ago

Explained precisely..fruitful return


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Prashanth Kumar R – over 2 years ago

Worked examples for each kind of possible problems would have been even more helpful. Do we have workedexamples for each of these formulas listed in the lesson?

If so, please provide me the link.

Thanks!


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LinkSuresh – over 2 years ago

Dear Prashanth,

Very soon we are going to come up with a series of lessons where all these concepts will be applied. I am sure

you will find them very useful

Regards


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Ishan Aggarwal – over 2 years ago

Gr8 stuff


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saka – over 2 years ago

Without Example or Test (indicating group like NUMBER OF PERMUTATIONS UNDER

CERTAIN CONDITIONS in hint )

This Material is loosing its significance.


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saka – over 2 years ago

CIRCULAR PERMUTATIONS::

There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a

circle so that there is exactly one person between the two brothers?

Hint: Point no 2 of circular permutations :-->

The number of circular permutations of n dissimilar things taken all at a time is (n-1)!

Solution:

If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and

this block of three people to be arranged around a circle.

The number of ways of arranging 18 objects around a circle is in 17! ways.********* -> 'n' objects can be

arranged around a circle in (n - 1)!.*****************************

Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways.

Therefore, the total number of ways 17! * 2 = 2 * 17!.


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Tushar – over 2 years ago

in the example given above i think the answer should be 17! * 18 * 2. Now you have choosen a block but we

need to choose it from either of 18 left so 18 ways to make that block.


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Dhara A. Mehta – over 2 years ago

good information sir…Can I know How much portion of this permutation & combinations will be asked in GREtest?


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Dear dharumehta07.

You need not get into too much of these concepts for GRE….just get to know about permutations,

combinations, circular permutations, seating arrangements, selection of members for committees etc;


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Rahul – over 2 years ago

A very good lesson to quickly revise all the formulae…Thank you Mr.Sureshbala


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venky – over 2 years ago

great stuff thank you very much , but examples for each of them would make it complete and awesome but still

great stuff for quick revision just before going to exam


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Dear coolvenky,

This lesson is created to revise the basic formulae keeping in mind the CAT 2008 test takers. I will definitrly try

to come up with series of lessons where all these concepts will be applied..


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ankit shrivastava – over 2 years ago

wonderful sites with great lessonssssss………..


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praveen yadav – about 1 year ago

ya …a great work bu the auther…..

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venky – over 2 years ago

sir how far is this topic important for the GMAT exam. i think i am weak in this topic, so i am worried.


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asureshwaran – over 2 years ago

this lesson is awesome sureshbala. you simply rock man.


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cash2323847 – over 2 years ago

pls solve this- There r 3 socialist,4 congressmen and 2 communist.In how many ways can a selection be made asto include at least one of each party.

Ans -315


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gargi_l – over 2 years ago

Number of permutations of n different things, taken all at a time, when m specified things always come together is

m

why not

m


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Oren Lahav – over 2 years ago

Say you have n items, and you're trying to organize them with m of the n items always coming together. (Clearlym < n). To organize the m items that come together, that's m!. Now to organize everything else, we have (n - m +

1)!, because we organize the items not in the M group, that's n - m, but we also have to put the M group together

with them, so in total it comes to n - m + 1.

I hope that makes sense.


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suman sourabh – over 2 years ago

good job


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– over 2 years ago

for someone like me who has v basic math knowledge examples of each type would be useful. few months ago

you said it was due - has it been created? I think we often get these types of questions in gmat


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– over 2 years ago

thanks


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anirudh92 – over 2 years ago

very good


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Vedatman – over 2 years ago

very good information……..thanks a lot sir


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Saurabh Roongta – over 2 years ago

awsome stuff ………..


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praveen yadav – about 1 year ago

i think its enough to prepare this ………


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greentree – about 1 year ago

not discriptive


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Linkkunal jain – about 1 year ago

i like this alot


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jatin luthra – about 1 year ago

thanx a lotbutdo u have similar notes for vectors n 3D too……………..


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itdoesntmatter – about 1 year ago

really nyc stuff


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amir saeed – about 1 year ago

please guide me how to solve this problem, if you joined all the vertices of heptagon, how many quadrilaterals willyou get?


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Linkharitham khan – about 1 year ago

although its a good rather best explanation but still iam not able to differentiate PC, so iam suggesting you to

differentiate them through examples and figures, u must use same example and figures for both then it will befriutfull


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Linkdevspring – about 1 year ago

thanks buddy its add on to ma skills


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Linkrnm_green – 6 months ago

nice representation


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abhishekfigo – about 1 month ago

Sir, I have a questions.

How many distinct circles can be made out of 5 points out of which 3 are collinear. Is there any general way tosolve it.

3 different points might have same circles.


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Name: Suresh

About: Worked for more than 6 years in renowned corporate institutes as their core faculty/lead content developer forC.A.T,G.R.E, G.M.A.T and Campus Recruitment Training Programs.

Posted Sep 11, 2008

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Permutations and Combinations Basics - CAT 2011