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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' waysIf 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 arrangementthat 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

63 Comments
    Sahanaz
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    SahanazSat, 31 Aug 2013 12:15:51 -0000

    how many different words can be formed from the letters a,c,d,e,z if vowels are never together and repetition is not allowed?

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    Sahanaz
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    SahanazSat, 31 Aug 2013 12:15:43 -0000

    how many different words can be formed from the letters a,c,d,e,z if vowels are never together and repetition is not allowed?

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    Sahanaz
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    SahanazSat, 31 Aug 2013 12:10:53 -0000

    how many different words can be formed from the letters a,c,d,e,z if two vowels are never together and repetition is not allowed ?

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    Rahulakki
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    RahulakkiThu, 29 Aug 2013 15:34:00 -0000
    D no. Of permutation of different things taken at a time which one particular thing never occurs

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    sagar_mgreen
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    sagar_mgreenWed, 29 May 2013 06:08:17 -0000

    Guys need help to solve this one.. How will we arrange Red balls in 6 places , so that if you choose any 3 consecutive places, there should be 2 Red balls among this 3 chosen places. (Number of Red Balls are not given)

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    rajesh2953
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    rajesh2953Tue, 14 May 2013 03:27:52 -0000

    The number of circles can be drawn out of 10 points of which 7 are collinear will be:

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    Mahmuddg
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    MahmuddgThu, 18 Apr 2013 10:10:34 -0000

    Combination a selection that can be formed by taking some or all a finite set of thing (or object) is called a combination

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    obyejekwu
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    obyejekwuMon, 04 Mar 2013 20:40:47 -0000

    Hi! Pls help me with this question. Six papers are set in an examination of which two are mathematical. In how many orders can the papaers be arranged so that: I) the two mathematical papers are together ii) the two mathematical papers are not consecutive

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    TALENTED BOY
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    TALENTED BOYTue, 19 Mar 2013 02:36:26 -0000

    The answer is (i) (n-1)! (ii) n!

    devdarguy
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    devdarguyThu, 21 Mar 2013 16:24:33 -0000

    how the second one is n!.

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    jben
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    jbenTue, 19 Feb 2013 07:59:40 -0000

    help me please!
    In how many different ways can 6 men and 4 women be seated in a straight line so that not two women are seated together?

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    Aleks
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    AleksThu, 25 Oct 2012 12:37:59 -0000

    Awesome explanation. It vud b gud if u give some examples 2 d related topic. If so plz post me those. :-)

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    Shuvajit
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    ShuvajitWed, 16 Jan 2013 20:49:44 -0000

    In a plane if there are n points out of which m points are collinear, then The number of polygons with k sides that can be formed by joining them is nCk-mCk. But what if m<k?? also why dont we subtract those polygons that are formed (hypothetically) by k-1 collinear points and 1 non-collinear point; k-2 collinear points and 2 non-collinear points;…….3 collinear points and k-3 non-collinear points???

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    rizwanqadirmemon
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    rizwanqadirmemonThu, 11 Oct 2012 16:20:03 -0000

    In how many ways 16 dollars can be divided into 4 beggars and no any beggar got less than 3 dollars.

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    Shuvajit
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    ShuvajitWed, 16 Jan 2013 20:58:19 -0000

    set aside 4*3=12 dollars from 16 dollars. so we are left with 4 dollars to distribute among 4 beggars. partition theory gives 7C3 ways of doing this.

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    AnupCool
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    AnupCoolThu, 20 Sep 2012 19:00:31 -0000

    These comments are really cool nd nice to revise

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    shiva235
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    shiva235Fri, 24 Aug 2012 12:33:42 -0000

    sir,what ever u kept before us will be very valuable….thank u

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    g0705
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    g0705Thu, 31 May 2012 19:45:34 -0000

    how many five digit numbers can be formed using the digits 0,2,3,4,and 5 when repition is allowed such that the number formed is divisible by 2 or 5 or both..plz solve this and post the answer plzzz.

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    udayprince299
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    udayprince299Sat, 02 Jun 2012 13:50:27 -0000

    numbers divisible by 2 are 1500
    numbers divisible by 5 are 1000
    numbers divisible by both 2 and 5 are 500

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    g0705
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    g0705Thu, 31 May 2012 19:24:49 -0000
    can any one plz solve this and tell me how to do it… In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S? thanks in advance..

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    gnv
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    gnvSun, 01 Jul 2012 03:54:58 -0000
    P and S can be arranged in 16 places;1 and 5 place,2 and 6 place,3 and 7 place ,4 and 8 place,5 and 9 place ,6 and 10 place,7 and 11 place,and 8 and 12 place since there should be 4 letters in between.Now we need to arrange remaining letters in between them.since there are 10 letters remaining ..10! ways is possible.but as there are 2 Ts repetition of counting is followed so divide it by 2! ….10*16
    fazimohd
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    fazimohdWed, 19 Sep 2012 04:33:55 -0000

    thr r 12 letrs in de word PERMUTATION. P nd S can b occupy in 1st nd 6th places/ 2nd nd 7th places/ 3rd nd 8th places/ 4th nd 9th/ 5th nd 10th/ 6th nd 11th/ 7th nd 12th places.. i.e., P nd S can b plcd in 7 ways. Also P nd S can be interchange their position. thus P nd S can be placed in 14 ways.
    since positions of P nd s are fixed, d remaining ten places in 10*14= 25401600

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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

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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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    g0705
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    g0705Fri, 01 Jun 2012 06:11:46 -0000

    its like choosins 4 vertices out of 7 vertices. coz quardilateral is of 4 vertices. now 7c4=7*4!=35..

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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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Last Updated At May 29, 2013
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