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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!=1×2x3x…..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).......

TOTAL NUMBER OF COMBINATIONS

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).

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)

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

Number of squares of any size in a square of n x n is \sum _{r=1} ^n r ^2

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}.

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Ask The Experts


  1. rajat_005 saidSun, 21 Sep 2008 14:25:12 -0000 ( Link )

    awesome stuff,really a lot of cocepts explained.

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  2. sanjucuckoo saidThu, 09 Oct 2008 16:44:42 -0000 ( Link )

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

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  3. Sureshbala saidFri, 10 Oct 2008 04:39:38 -0000 ( Link )

    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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  4. rdxb saidFri, 10 Oct 2008 18:29:51 -0000 ( Link )

    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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  5. pradeepdemi saidThu, 23 Oct 2008 10:14:40 -0000 ( Link )

    hi its very good i need probabilty with examples

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  6. ravi_shashank saidSun, 26 Oct 2008 11:27:31 -0000 ( Link )

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

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  7. vsagashe saidWed, 05 Nov 2008 08:59:18 -0000 ( Link )

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

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  8. swayam saidThu, 06 Nov 2008 14:50:34 -0000 ( Link )

    This is tremendously good yar very nice n easy explaination

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  9. aditi10 saidWed, 12 Nov 2008 19:08:52 -0000 ( Link )

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

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  10. hamsini saidThu, 20 Nov 2008 14:42:19 -0000 ( Link )

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

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  11. Sureshbala saidSat, 22 Nov 2008 07:43:06 -0000 ( Link )

    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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  12. santosh gupta saidSun, 30 Nov 2008 18:41:48 -0000 ( Link )

    u r gr8 sir

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  13. Sudharshan saidWed, 10 Dec 2008 23:32:42 -0000 ( Link )

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

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  14. pujii saidFri, 26 Dec 2008 16:47:21 -0000 ( Link )

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

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  15. prashanthpnu saidTue, 30 Dec 2008 15:44:51 -0000 ( Link )

    Very good lesson. Thank you :)

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  16. rajpatti saidFri, 02 Jan 2009 11:34:45 -0000 ( Link )

    excellent explanation on each topic great job thank u

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  17. abinash saidFri, 02 Jan 2009 16:10:03 -0000 ( Link )

    Explained precisely..fruitful return

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  18. prashanthpnu saidMon, 05 Jan 2009 17:40:37 -0000 ( Link )

    Worked examples for each kind of possible problems would have been even more helpful. Do we have worked examples for each of these formulas listed in the lesson? If so, please provide me the link. Thanks!

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  19. Sureshbala saidThu, 08 Jan 2009 07:45:29 -0000 ( Link )

    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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  20. Invincible Ishan saidFri, 16 Jan 2009 15:02:37 -0000 ( Link )

    Gr8 stuff

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  21. saka saidSat, 17 Jan 2009 05:52:18 -0000 ( Link )

    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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  22. saka saidSat, 17 Jan 2009 07:48:55 -0000 ( Link )

    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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  23. dharumehta07 saidSun, 25 Jan 2009 14:06:22 -0000 ( Link )

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

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  24. rahul2learn saidTue, 27 Jan 2009 12:55:16 -0000 ( Link )

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

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  25. coolvenky9 saidTue, 03 Feb 2009 09:44:17 -0000 ( Link )

    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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  26. shrivastavaankit2 saidThu, 05 Feb 2009 19:21:45 -0000 ( Link )

    wonderful sites with great lessonssssss…........

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  27. coolvenky9 saidSat, 21 Feb 2009 09:25:02 -0000 ( Link )

    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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  28. asureshwaran saidSun, 22 Feb 2009 16:17:49 -0000 ( Link )

    this lesson is awesome sureshbala. you simply rock man.

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  29. cash2323847 saidTue, 31 Mar 2009 16:29:35 -0000 ( Link )

    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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  30. gargi_l saidFri, 10 Apr 2009 13:27:07 -0000 ( Link )

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

    why not m!(n-m)!

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  31. spiyr saidSun, 19 Apr 2009 20:24:22 -0000 ( Link )

    good job

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  32. debjanir saidTue, 19 May 2009 13:50:49 -0000 ( Link )

    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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  33. debjanir saidTue, 19 May 2009 13:50:51 -0000 ( Link )

    thanks

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  34. anirudh92 saidSun, 24 May 2009 01:56:05 -0000 ( Link )

    very good

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  35. srinisv27 saidThu, 04 Jun 2009 00:28:31 -0000 ( Link )

    very good information….....thanks a lot sir

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  36. roongta_saurabh saidSun, 20 Sep 2009 09:12:27 -0000 ( Link )

    awsome stuff …........

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  37. prvnyadav51 saidSat, 17 Oct 2009 01:55:11 -0000 ( Link )

    i think its enough to prepare this …......

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  38. greentree saidWed, 09 Dec 2009 12:48:09 -0000 ( Link )

    not discriptive

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  39. kunal jain saidSun, 13 Dec 2009 07:05:26 -0000 ( Link )

    i like this alot

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  40. Swarn saidTue, 15 Dec 2009 21:06:08 -0000 ( Link )

    There are total 2n numbers among which three no.s are in A.P. Find the combinations of such three no.s from the total numbers? How to do it?

    Ans.:-  n(n-1)
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  41. jatin saidSat, 09 Jan 2010 18:47:39 -0000 ( Link )

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

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  42. itdoesntmatter saidSat, 09 Jan 2010 19:54:50 -0000 ( Link )

    really nyc stuff

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