Permutations and Combinations Basics

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’ ways

If an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then either ofthese two operations can be performed in ‘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.

PERMUTATION

An arrangement that can be formed by taking some or all of a finite set of things (or objects) 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 are arranged in a line. A linear permutation is simply called as a permutation.

A permutation is said to be a Circular Permutation if the objects are arranged in the form of a circle.

The number of (linear) permutations that can be formed by taking r things at a time from a set of n distinct things (r\underline<n) is denoted by ^n P_r \quad or \quad P(n,r).%

^n P_r = n(n-1)(n-2)(n-3)......(n-r+1) = \frac{n!}{(n-r)!}

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  r  (^{n-1} P_{r-1}).

2. Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement is  ^{n-1} P_r.

3. Number of permutations of n different things, taken all at a time, when m specified things always come together is  m!(n-m+1)!.

4. Number of permutations of n different things, taken all at a time, when m specified never come together is  n! - [m!(n-m+1)!].

5. The number of permutations of n dissimilar things taken r at a time when k(< r) particular things always occur is  [^{n-k}P_{r-k}] .[ ^rP_k].

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

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

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 n + n ^2 + n ^3 + ..........+ n ^r = \frac{n(n ^r -1)}{n-1}.

9. The number of permutations of n different things taken not more than r at a time  ^nP_1 + ^nP_2 + ^nP_3 +.....+ ^nP_r.

+ PERMUTATIONS OF SIMILAR THINGS+

The number of permutations of n things taken all tat a time when p of them are all alike and the rest are all different is \frac{n!}{p!}.

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 \frac{(p+q+r)!}{p!.q!.r!}.

CIRCULAR PERMUTATIONS

}1. The number of circular permutations of n dissimilar things taken r at a time is  \frac{^nP_r}{r}.

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

3. The number of circular permutations of n things taken r at a time in one direction is  \frac{ ^nP_r}{2r}.

4. The number of circular permutations of n dissimilar things in clock-wise direction = Number of permutations in anticlock-wise direction =  \frac{(n-1)!}{2}.

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  ^n C_r \quad or \quad C(n,r) \quad or \quad \tbinom{n}{r}.

1. ^n C_r = \frac{n!}{r!(n-r)!}

2. ^n C_r = ^n C_{r-1}

3. ^n C_r + ^n C_{r-1} = ^{n+1} C_r

4.If \quad  ^n C_r = ^n C_s \Rightarrow r = s \quad or \quad n=r+s

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

a)s particular things will always occur is  ^{n-s} C_{r-s}.

b)s particular things will never occur is  ^{n-s} C_r.

c)s particular things always occurs and p particular things never occur is  ^{n-p-s} C_{r-s}.

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  ^{m+n} C_m = \frac{(m+n)!}{m!n!}.

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  [\frac{(mn)!}{(n!) ^m}] \frac{1}{m!}

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  \frac{(mn)!}{(n!) ^m}.

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  \frac{(m+n+p)!}{m!n!p!}

5.The required number of ways of dividing 3n things into three groups of n each =\frac{1}{3!} \frac{(3n)!}{n!.n!.n!}.When the order of groups has importance then the required number of ways=\frac{(3n)!}{(n!) ^3}

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 (\underline< n) is  ^{n+r-1} C_{r-1}

}The number of non-negative integral solutions of the equation x_1 + x_2 + x_3+.....+ x_r = n \quad is \quad ^{n+r-1} C_{r-1}.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is  ^{n-1} C_{r-1}

The number of positive integral solutions of the equation x_1 + x_2 + x_3+.....+ x_r = n \quad is \quad ^{n-1} C_{r-1}.

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 x ^rin the expansion
(1 + x + x ^2 + ...........+ x ^p)(1 + x + x ^2 + ........+ x ^q).........

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 x ^r is the coefficient of x ^rin the expansion
(x + x ^2 + ...........+ x ^p)(x + x ^2 + ........+ x ^q).......

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

%{font-family:verdana}1.The total number of combinations of (p_1 + p_2 + ....+ p_k) things taken any number at a time when  p_1 things are alike of one kind,  p_2 things are alike of second kind…. p_kthings are alike of k ^{th} kind, is (p_1+1)(p_2+1).....(p_k+1).%

%{font-family:verdana}2.The total number of combinations of (p_1 + p_2 + ....+ p_k)things taken one or more at a time when  p_1 things are alike of one kind,  p_2 things are alike of second kind…. p_kthings are alike of k ^{th} kind, is%

(p_1+1)(p_2+1).....(p_k+1)-1.

SUM OF THE NUMBERS

Sum of the numbers formed by taking all the given n digits (excluding 0) is (Sum \quad of\quad all\quad the\quad n\quad digits)(n-1)!(111..n \quad times)

Sum of the numbers formed by taking all the given n digits (including 0) is (Sum \quad of \quad all \quad the \quad n \quad digits)(n-1)!(111..n \quad times)-

 (Sum\quad of \quad all \quad the \quad n \quad digits)(n-2)!(111..(n-1)\quad times)

Sum of all the r-digit numbers formed by taking the given n digits(excluding 0) is  (Sum \quad of \quad all \quad the \quad n \quad digits) ^{n-1}P_{r-1} (11111.......r \quad times)%

%{font-family:verdana}Sum of all the r-digit numbers formed by taking the given n digits(including 0) is  (Sum \quad of \quad all \quad the \quad n \quad digits) ^{n-1}P_{r-1} (11111.......r \quad times)- (Sum \quad of \quad all \quad the \quad n \quad digits) ^{n-2}P_{r-2} (11111.......(r-1) \quad times)

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  ^n P_r [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ...+ (-1) ^r \frac{1}{r!}].

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  n! [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ...+ (-1) ^n \frac{1}{n!}].

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  ^n C_2.

2. The number of triangles that can be formed by joining them is  ^n C_3.

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

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  ^n C_2 - ^m C_2 + 1.

2. The number of triangles that can be formed by joining them is  ^n C_3 - ^mC_3.

3. The number of polygons with k sides that can be formed by joining them is  ^n C_k - ^mC_k.

Number of rectangles of any size in a square of n x n is \sum _{r=1} ^n r ^3

In a rectangle of p x q (p < q) number of rectangles of any size is \frac{pq}{4} (p+1)(q+1)

In a rectangle of p x q (p < q) number of squares of any size is  \sum _{r=1} ^n (p+1-r)(q+1-r)

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  1 + \frac{n(n+1)}{2}.

Image Credits: cristic, cosmolallie, farouqtaj, churl

49 Comments
    softlaws4095 Vineet George
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    softlaws4095 Vineet GeorgeSun, 22 Apr 2012 06:35:15 -0000

    I have done extensive research on permutation and combination and found consistent result in my research. this result I have found is written in a book known as junction. To see my research result then log in to my sight

    https://sites.google.com/site/junctionslpresentation/home

    or e-mail me at Softlaws4095@gmail.com

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    reubenzibus
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    reubenzibusSun, 12 Feb 2012 17:00:35 -0000

    life has many struggles but only those that are determined win them… Mathematics needs determination beware!!!

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    Mathsmad
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    MathsmadFri, 10 Feb 2012 05:17:50 -0000

    hi….i hav a question sir. What r the types of questions r there in circular permutation? i need example for each type sir……..

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    mmnagpal
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    mmnagpalMon, 23 Jan 2012 10:35:55 -0000

    in short good knowledge to gain.

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    abhishekfigo
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    abhishekfigoMon, 22 Aug 2011 16:41:44 -0000

    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 to solve it.
    3 different points might have same circles.

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    rnm_green
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    rnm_greenFri, 04 Mar 2011 18:24:50 -0000

    nice representation

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    devspring
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    devspringSat, 14 Aug 2010 13:29:15 -0000

    thanks buddy its add on to ma skills

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    haritham khan
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    haritham khanFri, 11 Jun 2010 12:51:29 -0000

    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 be friutfull

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    amirsaeed
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    amir saeedSun, 14 Mar 2010 12:12:10 -0000

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

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    itdoesntmatter
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    itdoesntmatterSat, 09 Jan 2010 19:54:50 -0000

    really nyc stuff

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    jatin
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    jatin luthraSat, 09 Jan 2010 18:47:39 -0000

    thanx a lot
    but
    do u have similar notes for vectors n 3D too……………..

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    kunal jain
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    kunal jainSun, 13 Dec 2009 07:05:26 -0000

    i like this alot

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    greentree
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    greentreeWed, 09 Dec 2009 12:48:09 -0000

    not discriptive

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    prvnyadav51
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    praveen yadavSat, 17 Oct 2009 01:55:11 -0000

    i think its enough to prepare this ………

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    roongta_saurabh
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    Saurabh RoongtaSun, 20 Sep 2009 09:12:27 -0000

    awsome stuff ………..

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    srinisv27
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    VedatmanThu, 04 Jun 2009 00:28:31 -0000

    very good information……..thanks a lot sir

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    anirudh92
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    anirudh92Sun, 24 May 2009 01:56:05 -0000

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    debjanir
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    Tue, 19 May 2009 13:50:51 -0000

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    debjanir
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    Tue, 19 May 2009 13:50:49 -0000

    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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    spiyr
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    suman sourabhSun, 19 Apr 2009 20:24:22 -0000

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    gargi_l
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    gargi_lFri, 10 Apr 2009 13:27:07 -0000

    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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    oLahav
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    Oren LahavFri, 10 Apr 2009 13:47:54 -0000

    Say you have n items, and you're trying to organize them with m of the n items always coming together. (Clearly m < 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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    cash2323847
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    cash2323847Tue, 31 Mar 2009 16:29:35 -0000

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

    Ans -315

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    asureshwaran
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    asureshwaranSun, 22 Feb 2009 16:17:49 -0000

    this lesson is awesome sureshbala. you simply rock man.

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    coolvenky9
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    venkySat, 21 Feb 2009 09:25:02 -0000

    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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    shrivastavaankit2
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    ankit shrivastavaThu, 05 Feb 2009 19:21:45 -0000

    wonderful sites with great lessonssssss………..

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    prvnyadav51
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    praveen yadavSat, 17 Oct 2009 02:00:27 -0000

    ya …a great work bu the auther…..

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    coolvenky9
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    venkyTue, 03 Feb 2009 09:44:17 -0000

    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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    Sureshbala
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    SureshSun, 08 Feb 2009 20:50:59 -0000

    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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    rahul2learn
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    RahulTue, 27 Jan 2009 12:55:16 -0000

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

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    dharumehta07
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    Dhara A. MehtaSun, 25 Jan 2009 14:06:22 -0000

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

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    Sureshbala
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    SureshMon, 26 Jan 2009 20:51:41 -0000

    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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    saka
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    sakaSat, 17 Jan 2009 07:48:55 -0000

    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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    tushinform
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    TusharMon, 26 Jan 2009 20:07:49 -0000

    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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    saka
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    sakaSat, 17 Jan 2009 05:52:18 -0000

    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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About the Author

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

Last Updated At Apr 22, 2012
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