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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!=1×2×3x…..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 kind…. 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 kind…. 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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Suresh – over 2 years ago
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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Suresh – over 2 years ago
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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About the Author
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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