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Permutation and Combination

by Total Gadha - Sunday, 17 August 2008, 11:13 AM

Dagny has been telling me smugly that it's now my turn to write an article. So I decided to write one as soon as possible and pass the buck back to her. We both know that it takes a lot of pain to research for an article and write the text so we both try to avoid it for as long as possible. On the other hand, since we both enjoy writing and like to publish a well-finished chapter, we both undertake the task eagerly once the ball is in our court. Writing these chapters have made me realize how true is the maxim "if you want to learn something well, teach it." I can cite a lot of topics which I learnt while teaching in the class or answering students' questions. These years of teaching have made so many topics seem pretty childish to me which were nightmares to me before. But knowing a topic well and teaching it are two different ball games. Many a times, I have found myself struggling in the class to make a student understand something which I have found obvious to understand! The fault is not on the student's part. Three or four years ago, I would have been stuck on the same point! Every time I encounter a situation like this, my only measure is to stop at that point, retrace my steps with the students and slowly unravel the difficulty he's facing. It works most of the times. But sometimes, the student is in so much awe with the topic that his mind gets frozen. I have seen this happening many a times. Topics such as time, speed and distance, Permutation and C ombination, geometry etc. inspire so much fear among the students that sometimes simple principles, which they would have otherwise understood do not strike them as simple. I get incessant queries such as "Sir, time speed distance chhod sakte hain?," "Sir, permutation combination na karein to chalega?" from my students all the time. And the sad part is, that the level of C AT in these topics is very simple. They are doable. In the present chapter, I am trying to present one of these dreaded chapters as I understand it. I hope my students understand it too.

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I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this

in the CBT Club this week.

If you think this article was useful, help others by sharing it with your friends!

You might also like:

Probability

Maxima, Minima and Inequalities

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Re: Permutation and Combination

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Sir,

Tatz a GEM OF AN ARTICLE

thnk z,

VaMsI

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by Am it A - Sunday, 17 August 2008, 05:22 PM Now what do i say other than "THANKS A LOT TG !!! " u rock

....

Amit

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by Am it Kum ar - Sunday, 17 August 2008, 06:14 PM

Ve ry Ve ry use ful article .. Thank s

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by Le onidas S - Sunday, 17 August 2008, 06:20 PM

Hi TG,

A fabulous article .I think the re is a m istak e in the solution of Prob 3 of the 4 frie nds proble m . It says "The total num be r of ways in which 4 frie nds can stay toge the r is

104 " I think it should be 10000 as you found out e arlie r.

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by Am it Kum ar - Sunday, 17 August 2008, 09:40 PM Hi TG,

Answe r to 'Good Boy Bad Boy' m ovie que stion:

Since the se two couple s want to sit toge the r at e ithe r e nd. Now, we 'll tie the m toge the r. He nce , total num be r of stude nts will be 7 out of which 6 stude nts can sit in 6x 5x 4x 3x 2x 1 = 720 ways. and the couple can sit at e ithe r e nds (i.e 2 ways) and the couple s can sit in 2 ways am ong the m se lve s and finally, in e ach couple , boy and girl can sit in 2 ways(BG and GB)(Total = 2x 2 ways). He nce , Num be r of ways all can sit will be :

720x 2x 2x 2x 2 = 11520 (Answe r)

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by Manik a Tandon - Sunday, 17 August 2008, 10:41 PM

he y TG...

awe som e pie ce of writing dude ... rathe r thanx dagny for giving him a push to write it... :D both of u all rock ..

che e rs,

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by AsHwIn Drm z - Sunday, 17 August 2008, 11:02 PM hi....Tg....

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G8 article ...thk u so m uch for a wonde rful article ,actually i was worrie d as u we re not writing article s for cat 08 stude nts....but nw m so glad and confide nt dat m y all worrie s will vanish.Tg can u plz write an article on probability too....

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by Ashwin A - Monday, 18 August 2008, 07:51 AM TG Sir!

Sim ply Fantabulous!! Just wante d to k now do u have a Book having all topics of Q uant...I have the book on Num be r the ory which in itse lf,

is ve ry good.!!!

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by Dagny Taggart - Monday, 18 August 2008, 09:17 AM He y Manik a,

TG playe d sm art this tim e . I ask e d him to write an article thre e days back ;he wrote , e dite d and uploade d it within two days. Now the ball is back in m y court.

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by Diwak ar Bhask ar - Monday, 18 August 2008, 09:47 AM Thnk s a lot TG...

i dont have word to say anything m ore ...

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by vam si k rishna - Monday, 18 August 2008, 11:13 AM

He y Am it, Pl R e che ck

Shudn't Ans be 720 x 2 x 2 x 2

1..six stude nts can be arrange d in 6! = 720

2..The 2 couple s can sit at e ithe r of the e nds in 2 ways

3..and the n e ach couple on e ithe r side can e x change the ir se ats..in 2 ways so two couple s can do tat in 2 X 2 ways re gards,

VaMsI

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by Krushang Shah - Monday, 18 August 2008, 11:36 AM

THANKS A LO T

TG

!!! its a wonde rful article ...

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by am it k um ar - Monday, 18 August 2008, 01:44 PM

Thank s TG

Awe som e Article ...

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by nik hil k hande lwal - Monday, 18 August 2008, 04:58 PM Thank s a lot TG

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by rajde e p choudhuri - Monday, 18 August 2008, 05:46 PM

Hi TG,

The article is sim ply m ind blowing!!!Plzzzzzzzz post an e x e rcise on Pe rm utation and C om bination proble m s.

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by bryan cole - Tue sday, 19 August 2008, 12:59 AM

Is the answe r to a+2b+3c...total num be r of e le m e nts is 36?

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by Total Gadha - Tue sday, 19 August 2008, 03:37 AM Hi R ajde e p,

About the e x e rcise , have you se e n the article com ple te ly? Total Gadha

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by rajde e p choudhuri - Tue sday, 19 August 2008, 04:22 PM

Ye s .But i fe e l it will be lot m ore he lpful if you can give us an e x e rcise lik e the one on Num be r Syste m proble m s,I m e an ple nty m ore proble m s to solve .Also whe re can i ge t the answe rs to the se proble m s???

Anywys TG rock s!!!!

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by R iyaz Iqbal - W e dne sday, 20 August 2008, 11:03 PM Hi TG,

Ex ce lle nt article ,as usual.I've a que stion about the 'counting functions' proble m .The no.of functions from a 5-e le m e nt se t to 7-e le m e nt se t should be le ss than 7^5,right? 7^5 sim ply re pre se nts the num be r of ordered pairs (x ,y) from first se t to the se cond.Se ve ral of the m m ay corre spond to a single function.e g: (1,1) (2,2),...(5,5) are five distinct orde re d pairs but a single function viz y=x . Am I m issing som e thing?

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by ands ands - Thursday, 21 August 2008, 07:22 AM

ash, hi I've just starte d with the Num syste m but am finding the are a vast,e nom ously huge and the topics not conce ntrate d from a particular source .From your post I se e that the re is som e consolidate d re source for the Num syste m you talk in about or is the the e book by TG ppl.C an you post on the nam e he re .It would be sure he lp if I am to find the book and the conte nts satisfying.Thank s...!!!!

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by Total Gadha - Thursday, 21 August 2008, 01:18 PM Hi R iyaz,

The re are 75_{arrange m e nts, and e ve ry arrange m e nt, m apping 5 e le m e nts of se t 1 to 7 e le m e nts of se t 2 would be counte d as a single}

function only. Total Gadha

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by Shalini B - Thursday, 21 August 2008, 02:45 PM Hi TG,

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I have re giste re d for your C opycat e x am s by paying through cre dit card but I am facing proble m s while trying to write the e x am online . I have m aile d you long back about this (to the copycat e m ail ID and also to adm in e m ail ID). But I have not re ce ive d any answe r for that. And I could not find any contact num be r too on this site whe re I could call up and re port m y proble m . Ple ase re spond.

C onte nt from m y pre vious m ail date d July 30th 2008 to you be low:

My proble m is that whe n I trie d tak ing copycat 2 the site was unre sponsive and did not navigate to the ne x t page at all. I trie d tak ing the e x am but whe n I starte d with it and e nte re d m y options for the first page and click e d on the Ne x t button, the site be cam e ve ry slow and did not show the ne x t page at all. I trie d se ve ral tim e s but it was of no use . I che ck e d the browse r com patibility te st and it said m y syste m is fine for tak ing the te st and m y inte rne t conne ction is good and all othe r we bsite s are work ing fine . I am m ighty disappointe d. Ple ase look into this proble m ASAP

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by m anoj 1123 - Thursday, 21 August 2008, 04:06 PM Hi TG,

Though I fe e l not ok to write he re , I am also facing the sam e proble m , not able to browse to ne x t page . m ay be if the page re m ains ope n for som e tim e the conne ction to ur se rve r bre ak s.

Ple ase se nd the pdf for the C O PYC ATs to m y m ail be fore hand. So that I can tak e print and appe ar the te st..

Plz find a solution for us. Thank s.

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by Le onidas S - Friday, 22 August 2008, 10:33 PM Hi TG,

That was an e x ce lle nt article

C an you ple ase post som e article s on functions,logarithms and probability

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by rak e sh ojha - Sunday, 24 August 2008, 08:02 PM hi de ar

First of all ..thnx a lot for such a nice colle ctions of que stions which has he lpe d to build the fundam e ntals of this topic...

i just have one doubt in one of the e x am ple s ..above whe re we are ge tting num be r of 7 didgit num be rs ...in binary...that part is ok ... but whe n the num be r is conve rte d to de cim al base ....the re will be othe r conditions aprt from last 3 digits as ze ros...say a binary no whe n conve te d to de cim al and be com e s 24...that is divisble by 8....

m ay be i m wrong som e whe re but plz te ll m e why only you have tak e n the condition of last 3 didgits as 0 for dibisiblity by 8...why not othe r conditions....

thnx rak e sh

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by Total Gadha - Sunday, 24 August 2008, 10:12 PM Hi R ak e sh,

C onve rt 24 to binary ple ase ? Total Gadha

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by am it k um ar - Monday, 25 August 2008, 01:31 AM HI ALL,

que stion.>the re are 10 stude nts out of which thre e are boys and se ve n are girls,in how m any diffe re nt ways can the stude nts be paire d such that no pair consists of two boys?

Unable to solve this.Ple ase he lp wid prope r e x planation

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by IR O NMAN A - Monday, 25 August 2008, 11:57 AM

Thank s a m illion for the be autiful article TG. I have a sm all doubt. Plz do e x plain m e .

"There are 6 tasks & 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned either to person 3 or person 4. Every person must be assigned one task. In how many ways can the assignment be done(CA T 2006)?"

W hat's the wrong in m y proce dure ....

First person: As task 1 & task 2 cannot be assigne d any of the othe r 4 task s can be assigne d to him - 4 ways

Second Person: As task 1 & task 2 cannot be assigne d, and one alre ady assige nd to Pe rson 1, any of the othe r 3 task s can be assigne d to him - 3 ways

Third Person & Fourth person: one of the se ge t Task 2. The othe r ge ts any of the othe r 3. (one of task s 1,3,4,5,6 (le aving the task s assigne d to Pe rsons 1 & 2) - 3 ways

Fourth & Fifth Persons: othe r 2 task s can be assigne d in 2 ways. total = 4*3*3*2 = 72

Plz do answe r m e .. thank s in advance

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by arpi sin - Monday, 25 August 2008, 12:20 PM TG/Dagny

W onde rful work !!!

I was going through the num be r syste m quiz,and have a doubt in a que stion .W he re can I post m y que ry ? R e gards

Arpita

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by Sure ndran C handravathanan - Monday, 25 August 2008, 09:43 PM Hi Am it,

He re 's m y try... 1st A pproach

1 boy can pair up wid 7 girls in 7 ways...So for 3 boys, total no. of ways is 21..And 1 girl can pair up wid 6 othe r girls in n(n-1)/2, ie , (7x 6)/2 = 21 ways..(or) Just 7C 2 = 21 will ce de u the answe r.

So the final ans is 21 + 21 = 42 ways.. [[O R ]]

2nd A pproach

Should this proble m be solve d lik e be low....???

To se le ct 2 ppl from 10, we can do in 10C 2 = 45 ways..And arranging the m in pairs can be done in 45x 2 = 90 ways..(or) juz we have to arrange 2 place s out of 10 which can be done in 10P2 = 90 ways..

And the n, 3 boys can be paire d up in 6 ways.. So is the final ans 90 - 6 = 84 ways ?????? I am pe rple x e d..

Which ans is correct, TG ????? A ppreciate ur comments... Be st R gds,

Sure n

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by Sure ndran C handravathanan - Monday, 25 August 2008, 09:58 PM

TGGGGGG,

Ur art of solving & sim plifying the proble m s le ave s m e in awe but it invite s m ore confusions... I wud say dat the be low proble m can be solve d e asily but u said it is not so..

In how many ways can four persons be seated out of 5 boys and 3 girls on four different seats? Now se e , totally thr are 8 ppl whe re the y 've to be se ate d in 4 se ats..

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ie , 8P4 = 1680 ways....

Plz e lucidate the fact for bre ak ing the proble m & conside ring 4 case s in ur approach.... Thank s in advance ,

Be st R gds, Sure n

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by Black i - Monday, 25 August 2008, 11:38 PM Hi TG,

It is re garding the '7 digit binary no div. by 8 when converted to decimal no' que stion.

How can one de cide that the last 4 bits , if are e qual to ze ro the n the no is divisible by 8 whe n conve rte d to de cim al ? Suppose the sam e que stion ask s about octal no inste ad of binary, how should one de cide about it ?

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by am it k um ar - Tue sday, 26 August 2008, 10:21 PM Hi

Answe r is 630.give anothe r try.

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by Harish Bansal - Thursday, 28 August 2008, 12:51 PM

Hi TG,

Gr8 article....

Plz post the solutions of the problems that you have mentioned in ur artical...

We want to match our answers and approach.

And plz can u post something on probability prior to CAT 2008.

Eagrly waiting for a reply...

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by Tuhin Bane rje e - Tue sday, 2 Se pte m be r 2008, 05:14 PM

Hi TG, it's realy a mind blowing article on permutation and combination.

I have question on the same topic and have two different approaches , kindly tell me which one correct :

Question: Ten students have been shortlisted to form two teams of six students each, such that there are exactly three common

between the two teams, In how many ways can the teams be formed.

Ans:

1st approach:

Three common stuednts we can take out of 10 students =

10

C

3

ways

,

We need to build each group having 6 students , for the 1stGR. rest of the 3 students we can take out of 7 students =

7

C

₃

ways

So , 1stGR complete, for the 2ndGR we can take rest of the 3 students out of the 4 students in =

4

C

₃

ways.

Total =

10

C

3

*

7

C

3

*

4

C

3

= 16800

2nd Approach :

From 10 students , 3 students commonbetween the two teams can be selected in =

10

C

₃

ways.

since 7 students are still left and for each team we need 6 students , Now select the student

will find a place in neither team and it can be done in =

7

C

₁

ways.

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Now remaining 6 students have to divided into two groups of 3 each and it can be done in

=

6

C

3

/ 2

Now two teams can be formed =

10

C

3

*

7

C

1

*

6

C

3

/ 2 = 8400

please tell me where I am missing the concepts

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by rudra sam al - Tue sday, 2 Se pte m be r 2008, 10:01 PM Hi TG,

In the e x am ple whe re u ask e d:

How m any arrange m e nts r de re of six 0's, five 1's and four 2's, whe re (i)first 0 pre ce de s first 1?

shudnt the ans be 343980 and not 349380?.... and the re 's an e asie r way as we ll. It goe s lik e this: total arrange m e nts=(15!/(4!*5!*6!))

so ans will be =total arrange m e nts*(p/q) whe re q=all possibl arrngm nts of the 1s and 0s p=arrngm nts whe re 0 occupie s 1st position so,

q=11!/(6!*5!) p=10!/(5!*5!) ans=343980

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by Total Gadha - W e dne sday, 3 Se pte m be r 2008, 10:49 AM Hi R udra,

C orre cte d the typing e rror. And good way to solve the que stion. Total Gadha

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by Total Gadha - W e dne sday, 3 Se pte m be r 2008, 10:51 AM Hi Tuhin,

The se cond solution is corre ct. First has re pe tition. Ple ase che ck the que stion of "thre e state s with thre e stude nts from e ach state ...." Total Gadha

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by bhage e rath n - W e dne sday, 3 Se pte m be r 2008, 10:30 PM hi

i'm he re for the first tim e .. this was pre tty he lpful.. how do i cross che ck if i got the right answe rs?? will the solutions be available for the que stions??

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by bhage e rath n - W e dne sday, 3 Se pte m be r 2008, 11:04 PM Hi TG,

for the que stion

-In how m any ways can four pe rsons be se ate d out of five boys and thre e girls on 4 diffe re nt se ats? should we tak e those case s as m e ntione d by you??

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the re are 8 pe ople in all.. se le ct four of the m in 8c4 ways (70 ways). Now you can arrange the se four in 4! ways(24 ways). So total ways is 70*24 = 1680.

is the re anything wrong with this approach??

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by Syd S - Thursday, 4 Se pte m be r 2008, 01:58 PM

k udos for anothe r brilliantly structure d and e dite d article ,,,!!!! was just wonde ring guys bout dis prob,,,

"in how m any diff ways can the face s of a cube be painte d in 6 diff colors?" shudnt the no of ways be 6 X 5 x 3!,,

as the first face can be painted in 6 ways,, rathe r than only 1 way,,

jus wante d 2 k nw,,y is it 5 X 3! rathe r than 6 X 5 x 3! thnx

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by brije sh gogia - Saturday, 6 Se pte m be r 2008, 07:12 PM Thank You for this ve ry use ful article .

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by gane sh p - Sunday, 7 Se pte m be r 2008, 08:41 AM

That U Ve ry m uch sir ...as e x pe cte d m y se arch for the be st article on pe rm utation and com bination did e nd he re ..

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by k ladad asda - Sunday, 7 Se pte m be r 2008, 08:44 PM

just be ing curious. are u vam se e k rishna from C SE IITKGP, class 2006??

If you are k indly re ply back , it would be nice to k now one m ore fam iliar face at TG

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by Manish Kashyap - Tue sday, 9 Se pte m be r 2008, 07:56 PM

Hi TG,

My Solutions are

-1)a

2)c

3)b

4)

5)c

6)

7)c

8)c

9)a

10)b

11)b

12)c

13)b

14)b

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15)

16)

17)b

19)d

20)

21)c

22)16725

23)24

24)840

25)120

26) 680

27)26!/2

28)45

29)21

30)35

31)5

32)7c2

33)33600

I was not able to solve some questions for which I had not put any answers. Can you please verify my answers and

tell me the solutions of which are wrong or which I was not able to solve.

Thanks and Regards,

Manish

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by Ank it Arora - W e dne sday, 10 Se pte m be r 2008, 05:27 PM Hi TG,

I have n't solve d all the que stions till now...just want to confirm the answe rs of first 5 que s.. 1 (a)

2(b) 3(d)

4. I m ge tting (26^3)*220 as m y answe r 5(c)

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by Manish Kashyap - Thursday, 11 Se pte m be r 2008, 03:09 PM

Hi TG Sir,

Ple ase che ck the solutions ple azzzzzzzzzzzzzzz. I am waiting for solutions. Thank s and R e gards,

Manish

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by vam si k rishna - Friday, 12 Se pte m be r 2008, 04:25 PM Frns

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Check/discuss answers for above problems @

http://totalgadha.com/mod/forum/discuss.php?d=3683

VaMsI

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by Saurabh She k har - Friday, 26 Se pte m be r 2008, 09:59 AM Hi TG,

C ould you ple ase post the com ple te solution to the Ex e rcise . W e are all e age rly waiting for it.

Saurabh

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by Ve nk k ate san R - Saturday, 27 Se pte m be r 2008, 04:44 AM Hi sure n,

I think the que stion involve s just the se le ction. He nce the arrange m e nt is not re quire d. He nce 10c2-3c2=45-3=42 is the answe r.

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by arvind je nam ani - Saturday, 4 O ctobe r 2008, 09:19 PM

Hi TG sir,

The link is fabulous. In com bination part the re is a que stion " In how m any ways 5 boys , 3 girls can be sitte d in 4 se ats?" The re we don't ne e d to se gre gate into parts lik e you m e ntione d.

8C 4 ways we can se le ct 4 pe ople (without conside ring boy,girl com bination) 4! ways to arrange the m . so, 8C 4*4!=8P4=1680

which m atch also your answe r.

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by Am it Trive di - Saturday, 11 O ctobe r 2008, 10:31 PM

Hi TG sir,

The lesson is really great. i have a doubt for one of the examples.

the examplein which 5 boys and 5girls go to watch the cinema spiderman3.

you gave two arrangement for the wish of the boys "every boy wants to sit with a girl" and no two girls sit together

i think other arrangement of the form are possible

G B B G B G B G B G

G B G B B G B G B G

G B G B G B B G B G

G B G B G B G B B G

In the above arrangements every boy is sitting with a girl and no two girls are sitting together

clarify my doubt, let me know where i am making mistake in reasoning.

Warm Regards,

Amit

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by Am it Trive di - Saturday, 11 O ctobe r 2008, 11:57 PM if the couple s are siting at the e nds

the arange m e nts are 2 X 2 X 2 x 6! Am it

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by m e e na k - Saturday, 18 O ctobe r 2008, 04:24 AM

hi,

for the states question can't we solve like this from 3states 3 students eachso selecting 5 frm 9= 9c5 then subtracting 6c5 (case where all 5 students are selected frm two states)

ie:9c5 - 6c5 .pls tell me where the problm is

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by noe l sande sh - Monday, 3 Nove m be r 2008, 03:40 PM Hi TG,

This is in re gard to an e x am ple proble m (No. of 7 digit binary num be rs that are divisible by 8 whe n conve rte d to base 10). i was wonde ring as to y the 3rd digit from the last cannot be 1 or 0. You've m e ntione d that the last 3 digits should be 0. But if the last 3 digits are 100 the n its base 10 e quivale nt will be 400 which is divisible by 8.... I'm re ally bad at P n C but i can't figure out the logical flaw in m y thought... C an u he lp m e out on this?

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by raje sh k am bam pati - Thursday, 6 Nove m be r 2008, 12:36 AM

Ya i agre e with you sande sh, i too got this doubt.And i think we are corre ct.

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by Total Gadha - Friday, 7 Nove m be r 2008, 12:49 AM (100)2 = 4 (in de cim al)

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by C om ple te Gadha - Friday, 7 Nove m be r 2008, 08:03 PM Hi TG Sir,

This is the best material i have ever come across for Permutation and Combination.

I have been going through the material in the forum. all of them r so amazing that i am actually gaining the confidence to crack the CAT though i have not been doing so good in the mock cat. Thanks a lot..!!

Hats off To U...!!!

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by Bhe e m e sh K - W e dne sday, 12 Nove m be r 2008, 01:40 AM Hi TG,

I had a doubt in the first proble m (i.e 10 spe ak e rs proble m ) solve d in the pe rm utation se ction.

I solve d it in this m anne r. you tie the PM, MP and MLA into a single pe rson and the n calculate the ways to arrange the m . wudn`t it be 8! the n that we have othe r 7 spe ak e rs and 1 group of PM, MP and MLA to arrange ?

I find your way also logical but am not able to unde rstand diffe re nce be twe e n m y approach and your approach. k indly can you e x plain whe re I am m issing the logic?

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by sum it jam wal - W e dne sday, 4 Fe bruary 2009, 08:51 AM

Hi TG,

I have som e proble m in unde rstanding the conce pt of partition proble m .. of 5 balls as state d be you.

le t's tak e * * * * * the se as 5 sim ilar balls now dividing the m into 3 grps..

i can do it by placing 2 stick s ..for that i have total of |*|*|*|*|*| --> | de note s choice s

6 choice s..

so se le cting 2 space s out of 6 can be done in 6C 2 = 15 ways..

te l m e wats goin wrong in this approach..

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by sum it jam wal - Thursday, 12 Fe bruary 2009, 09:41 PM hi TG,

the proble m i had above , i'd done a m istak e actually thr wud be 30 ways ...

for 1st stick - 6 ways for 2nd stick - 5 so total - 6*5=30.

if you cud clarify this thing it wud be nice R e gards,

sum it

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by Total Gadha - Thursday, 12 Fe bruary 2009, 11:09 PM Hi Sum it,

In your solution, you can't place stick s lik e this for e x am ple ||* * * * *

This would de note a 0 0 5 case which you cannot cove r. You can only cove r 0 5 0 case by placing the stick s in this m anne r: |* * * * *| The way you are placing the stick s you can ne ve r place two stick s side by side be cause you are tak ing it to be one space only. Total Gadha

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by sum it jam wal - Friday, 13 Fe bruary 2009, 12:16 AM hi sir,

sir that's fine ..bt if iam nt conside ring this case ..the n y is m y answe r com in to be arnd 30 ..whr as u state d it's 21. i think thr m ight be som e k ind of re pe atition...

and as u told

| | * * * * * nt include d so, now suppose i place 1st stick the n arrange m e nt wud be lik e .. | * * * * *

now to include ur case for 2nd stick thr are 6 case s as i can place stick be twe e n stars and be side 1st stick

in this case total case s com e out to be 36 still m ore than 21 a bit confuse d as how to ge t corre ct re sult by using this stick m e thod.

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re gards, sum it

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by Total Gadha - Friday, 13 Fe bruary 2009, 09:32 AM Hi Sum it,

The re pe tition is be cause you are tak ing partitions to be diffe re nt whe re as the re have to be tak e n as sim ilar. For e x am ple , tak e two sim ple case s:

* P1 * * * P2 * and * P2 * * * P1 * whe re P1 and P2 de note the partitions. Both the se case would yie ld the sam e re sult, i.e . 1 3 1 but you

are counting the m as diffe re nt.

If I do this by your m e thod I will do it lik e this: For the first partition I have 6 place s.

* * | * * *

For the se cond partition I have 7 place s. The re fore total num be r of ways = 6 × 7 = 42. As e ach case is be ing re pe ate d once , the total num be r of case s = 42/2 = 21.

Total Gadha

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by sum it jam wal - Friday, 13 Fe bruary 2009, 05:07 PM

hi TG,

thank s ,that was wat i think ing....

gr8 work .. R e gards , sum it

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by Abhinav Kum ar - Friday, 22 May 2009, 01:38 PM

Thats re ally a be autiful one .

Made such a difficult task so sim ple with the e x am ple s of such a varie ty. Thank s a Lot

Abnv

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by rony roy - Friday, 22 May 2009, 02:33 PM hi TG,

what was the answe r to the que stion that goe s...

how m any arrange m e nts of six 0s five 1s and four 2s are the re in which ii)the first 0 pre ce de s the first 1,(and) pre ce de s the first 2.

m y answe r com e s out to be a m assive : 504504

re ply soon...

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hi tg.. 1st of all thank s a ton for this sim ply awe som e post .. i ws nt ge tting any gud m ate rial on P&C n Probability n was worrie d ... thank s again..

.. re garding the Problem :: the re are 10 spe ak e rs who r suppose d 2 addre ss. O ne constraint is PM shud addre ss be fore MP and MP be fore MLA .

My soln:: Le t PM = P ; MP = Q ; MLA = R

the orde r shud be PQ R .. so i tuk PQ R as 1 block and so total pe ople 2 b arrange d be com e s (7+1) = 8 ... and now arranging the m we ge t A ns = 8! .

Ple ase te ll whe re am i m ak ing the m istak e ?

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by Priye sh Tungare - W e dne sday, 27 May 2009, 02:59 PM Hi TG,

Thank s for the article . I have one doubt.

In the que stion, How m any 3-digit num be rs are e ve n and with no re pe ate d digits, in the se cond case for num be r not e nding in 0, you have m ultiplie d 8*8*4.

C an you e x plain why we are m ultiplying 4 he re ?

This m ay be sim ple que stion for othe rs, bt i be lie ve in saying.. whe n in doubt ask .... Thank s,

Priye sh Tungare

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by Total Gadha - Thursday, 28 May 2009, 02:06 PM Hi R ande e p,

'Be fore ' m ight not always m e an 'just be fore ,' as you have assum e d by tak ing the m toge the r. The re can be othe r pe rsons be twe e n P, Q and R .

Total Gadha

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by Total Gadha - Thursday, 28 May 2009, 02:08 PM Hi Priye sh,

The re are four 4 e ve n digits afte r re m oving ze ro- 2, 4, 6 and 8. The num be r would e nd in one of the se . Total Gadha

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by iim fre ak - Thursday, 28 May 2009, 10:21 PM

oh ye ah ,, gosh i shud hav thot dat.. thank s a lot for cle aring the doubt ...

TG sir it will b a gr8 he lp if u cud post som e gud te x t, lik e dis, for Probablity also .. dats anothe r are a, which i find tough..

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by iim fre ak - Friday, 29 May 2009, 10:09 AM

hi tg.. from whe re can i find the answe rs 2 the above e x e rcise .. its v im p 2 chk whe the r i hav unde rstud ur conce pts pr nt.

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by anirban bhar - Sunday, 31 May 2009, 12:34 PM

hi tg,

im a bit confse d wid the solution to the paint cube proble m ...

the firstface wich we choose cn b painte d in 6 ways,...y have nt u conside re d that....plzzz e nlighte n

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by Avishe k C hak raborty - Thursday, 11 June 2009, 07:39 PM

He y guys I have a que stion re late d to this topic, could u solve this? He re it goe s...

Q . 4 boyz n 4 girls are arrange d in a row so that no 2 girls are toge the r.W at is the probability that any 2 boyz are toge the r? waiting for ur re sponse ...

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by iim fre ak - Thursday, 11 June 2009, 08:27 PM

dude .. this is PnC thre ad n u ask ing Probability ..jst k idding .. anyways i dnt k now probability so cant he lp @ this m om e nt .. TG sir will be posting a te x t on Prob soon... waiting for that...ATB m ate

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by k u k lux k lan - Thursday, 18 June 2009, 07:18 PM the re are 4 frie nds and 10 hote l room s

q)in how m any ways can at le ast two frie nds stay toge the r

m y answe r to this is as follows,which is le ading m e to the wrong answe r,will be ve ry grate ful if som e one can point out the fallacy in this 1 case )-2 frie nds stay toge the r

c(4,2)*10*9*8

2nd case )thre e frie nds stay toge the r c(4,3)*10*9

3rd case )all four stay toge the r 10

adding up all the se case s i am ge tting 4690,which is wrong....ple ase te ll m e the m istak e in this

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by rohan k aushal - Saturday, 20 June 2009, 09:03 AM hi,

Q . how m any thre e digit num be rs are the re which are e ve n and have no re pe ate d digits ????

can't it be this way??? _ _ _

the 3rd place 5 choice s ,(i.e ; 0,2,4,6,8) so no of ways are 5 1st place have 8 choice s (e x cluding 0 and the 3rd place 's num be r) & 2nd place have again e ight choice s ....

and that com e s out to be 8 * 8 * 5 = 320 .. could you plz he lp m e ...

thanx a lot!!!!

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by am it jain - Sunday, 21 June 2009, 12:37 PM

@R ohan,

W he n you will use 0 for the unit's place the n for the hundre ds place you will have 9 choice s(1....9) and 1oth place 8 choice s. I hope now you k now whe re you are ge tting it wrong

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by srinivasan ravi - Sunday, 12 July 2009, 10:51 AM hi sir,

can u pls e x plain m e the proble m just be fore cicular com binations..i cant unde rstand it cle arly..pls he lp..thank s..

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by sm ruty panigrahi - Monday, 13 July 2009, 11:10 AM

Hii,

This is a rathe r sim ple look ing proble m but has be e n pe ste ring m e and am unable to solve . Pl che ch it out and he lp m e guys. You have to find out the num be r of ways to re ach from A to B.

Ex tre m e ly sorry 4 the poor quality of fig. I ain't that good with com pute rs. Plz give a de taile d solution.

Thanx in advance .

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by aryan raj - Monday, 13 July 2009, 03:03 PM

hi sir

this file is so good for cat e x am so i want solution of the all prob so plz sand m e the solution on m y e m ail

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by Ank it Me gotia - Tue sday, 28 July 2009, 11:13 AM he y sm ruty,

(29)

its com ing out to be 72.

if its corre ct i will te ll you the proce dure .

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by Jaide e p Das - Tue sday, 28 July 2009, 10:44 PM

C an som e body provide m e the answe rs for (ii) and (iii) parts of the que stion for 6 0's, 5 1's ans 4 2's? I got the answe rs as 15840 for (ii) and 378738 for (iii)

Are the se answe rs corre ct? Ple ase re ply

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by Gowtham Muthuk k um aran Thirunavuk k arasu - Monday, 3 August 2009, 11:20 PM You r right.

The solution give n he re , is base d on a assum ption that the thre e spe e k in orde r, which is not the case . Your solution is pe rfe ct i gue ss.

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Re: Permutation and Combination to Bheemesh K

by Gowtham Muthuk k um aran Thirunavuk k arasu - Monday, 3 August 2009, 11:22 PM You r right.

The solution give n he re , is base d on a assum ption that the thre e spe e k in orde r, which is not the case . Your solution is pe rfe ct i gue ss.

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Re: Permutation and Combination TO smruty panigrahi

by Gowtham Muthuk k um aran Thirunavuk k arasu - Tue sday, 4 August 2009, 12:31 AM

From the figure you have give n i have work e d out and got the answe r to be 30. Juz che ck if you alre ady k now the answe r and if it is wrong ple ase le t m e k now. I have attache d the file he re .

More ove r I have assum e d that one can trave l through a path only once . For e x am ple you cannot go through ED and again com e back to E.

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by Avishe k C hak raborty - W e dne sday, 5 August 2009, 03:31 PM

Firstly i m ust say the article is awe som e ! Ple ase pardon m e if m y que stion se e m s childish!

The re 's one que stion re late d to dice , in how m any ways sum of 8 can be obtaine d by rolling 2 dice if the the y are distinguishable ... now m y que stion is as we are rightly conside ring (3,5) and (5,3) as diffe re nt case s, why are we not tak ing two case s for (4,4)? Though the de nom inations are sam e but the dice are distinguishable .

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by cat cham p - Saturday, 8 August 2009, 04:11 PM

i dont think it will be tak e n abhishe k .. as(4,4) is m ak ing the dice indistinguishable ..

btw.. ve ry good work tg sir..the be st article till date i have found on p&c.. and i will be posting the solns soon..

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by anim e sh chandan - Saturday, 15 August 2009, 10:57 PM hi TG

i hav proble m

Q :the re are 7 m e n and 5 wom e n we have to se le ct 4 m e n and 3 wom e n but if m rs A will go the n m rs B will not go.in how m any ways v can se le ct the m ?

m y approach is 7 m e n will be se le cte d in 7C 4 ways and the re are 3 case s

case 1: we first se le ct m rs A the m autom atically m rs b will not be se le cte d the n we can se le ct re m aining wom e n in 3C 2 ways, that is 7C 4*3C 2=105.

case 2: we se le ct m rs B the n autom atically m rs A will not be se le cte d the n ve can se le ct re m aining wom e n in 3C 2 ways, that is 7C 4*3C 2=105,

case 3:if both of the m are not se le cte d the n we will have 3C 3 ways to se le ct wom e n that is 7C 4*3C 3=35

that m e ans total no of ways =105+105+35=245 plZZZ te l m e is it the right approach?????

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by Anoop Maithani - Thursday, 20 August 2009, 10:30 AM Hi TG..

Am azing article ..waiting e age rly for an article on Probability..

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by Be llThe C AT ... - Thursday, 20 August 2009, 07:18 PM

Hi TG,

A re ally he lpful article ,for it he lpe d m e re vise P&C . thank s a lot.

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by de e p sin - Saturday, 5 Se pte m be r 2009, 01:59 PM @ k u k lux k lan

You are forge tting to tak e one case into calculation

The case is : - 2 stay in sam e hote l and re m aining 2 also stays in sam e hote l.

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by Pallav Jain - Thursday, 17 Se pte m be r 2009, 07:03 PM

Hi All,

C an you plz te ll m e the solution ?

(31)

O ptions

:-A) 75_{+ 7}4_{+ 7}3_{+ 7}2_{+ 7 C ) 4 × 7}6

B) 4 × 75 D) ₇6 E) _{6 × 7}6 Thanks

Pallav Jain

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by priya ram raj - Friday, 18 Se pte m be r 2009, 04:00 PM Hi TG,

I have a doubt in the article that u have poste d. To ge t a sum of 8 with distinguishable dice s, wont the occurre nce of(4,4) be diffe re nt be cause it is distinguishable ??? shudnt we conside r it as 2 case s ..Pls clarify...

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by R O HIT K - Friday, 18 Se pte m be r 2009, 05:05 PM Hi priya,

Jus to giv it a try..

To get a sum of 8 with distinguishable dices, wont the occurrence of(4,4) be different because it is distinguishable??? shudnt we consider it as 2 cases ..

(2,1) and (1,2) are diffe re nt in distinguishable dice (which is not the case in sam e k ind of dice ). Say, you have a re d and a blue die

C ase : (1,2) and (2,1) 1.R e d->1 Blue ->2 2.R e d->2 Blue ->1

You can distinguish the two not only by the ir color but also by the diffe re nce in the ir num e rical value . C ase 2: (4,4)

1.Blue ->4 R e d->4 2.R e d->4 Blue ->4 Are n't the y both sam e ??

If ye s, the n we ne e d not conside r it two case s.

In short, since the num e rical value is sam e , we cannot distinguish the case though the two dice are distinguishable . Hope this he lps.

R ohit

P.S if you still didn't unde rstand...TG to hai..!!

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by priya ram raj - Friday, 18 Se pte m be r 2009, 05:31 PM Hi R ohit,

going by ur e x am ple ,W at i unde rstand from the article is that

color is the distinguishing factor and not the num be r... If we re m ove the color aspe ct from the argum e nt only the n the distinguishing prope rty vanishe s... hope i am not wrong..anyways.. Thank s for the approach

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by R O HIT K - Saturday, 19 Se pte m be r 2009, 02:17 PM Hi priya,

Color is the distinguishing factor and not the number...

If we remove the color aspect from the argument only then the distinguishing property vanishes... I do not com ple te ly agre e with your conclusion..

The dice had two diffe re nt colors viz. re d and blue ..(I did not re m ove the m )

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The re ason was not color but sam e num e rical value ..

If you had change d the num be r to <1,2> the n you would have distinguishe d it from <2,1> occurre nce . if you can che ck m y e x planation again afte r re ading this hope it would be m ore cle ar..

1. If colors of dice are sam e (i.e indistinguishable dice ) the n occurre nce of <1,2> or <2,1> or for that m atte r any pair of value s and vice ve rsa cannot be distinguishe d..(ve ry obvious) (indistinguishable be cause of color)

2. If colors of dice are diffe re nt the n

a. <1,2> can be distinguishe d from occurre nce of <2,1>(distinguishable be cause of color of the dice )

b. <4,4> or for that m atte r any sam e value on both dice cannot be distinguishe d from vice ve rsa value (which is actually the sam e occurre nce as the pre vious..) (indistinguishable be cause of sam e value on the dice )

He nce , the conclusion can be drawn that the distinguishing factor in the dice of diffe re nt colors is the color aspe ct, only whe n num be rs appe aring on the dice are diffe re nt. If the num be r be com e s sam e (indistinguishable ) the n you cannot distinguish the occurre nce . Hope this he lps..

R ohit

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by Sure sh S - Monday, 21 Se pte m be r 2009, 11:11 AM Hi sir ..

its re ally supe rb.... thank s for ur article ...

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by R O HIT K - Tue sday, 22 Se pte m be r 2009, 12:36 PM Hi TG..!!

Thank s for such a nice article . I had one que stion re garding:

In how m any ways can 4 pe rsons be se ate d out of 5 boys and 3 girls in 4 diffe re nt se ats? W hile e x plaining you said it is not the sim ple form ula base d P&C proble m .

Though, I think that it could be solve d in ve ry sim ple te rm s without bre ak ing down the proble m into case s(the way you alre ady sugge ste d as the conce pt).

Two aspe cts to this proble m :

1. Se le cting 4 pe rsons from 8 pe rsons(5 boys+3 girls) which can be done in 8C 4 ways

2. Arranging the 4 pe rsons on 4 diffe re nt se ats for e ach se le ction which can be done in 4! ways So total num be r of ways = 8C 4 * 4! = 70*24 = 1680 (sam e as 8P4)

Ple ase le t m e k now if this is not the right way to approach the proble m . I ne e d your insight into this.

Thank s and R e gards R ohit

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by R ohan S - Tue sday, 29 Se pte m be r 2009, 11:49 AM

TG STANDS FO R TR UE GUIDE...

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by sharm ishtha Gupta - Friday, 9 O ctobe r 2009, 01:13 PM Hi,

supe rb article . i have a doubt though,

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can we first m ak e the boys sit in 5! ways _B_B_B_B_B_

now, for the girls to sit , we have to choose 5 of the 6 place s available so that no two girls sit toge the r, this can be done in 6 ways and the girls can the n be se ate d in 6* 5! ways.

The re fore , the no of ways in which the y can sit according to the give n condition the n com e s out to be 6*5!*5!.

Ple ase corre ct m e if i am wrong. Thanx

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by AsHwIn Drm z - Saturday, 10 O ctobe r 2009, 01:32 AM hi Sharm ista,

You are m issing out som e hig he re . Say 1st we se at all the boys..in 5! ways.

The n we have 6 place s whe re 5 girls have to be se ate d.Now, 1st girl can be se ate d in any of the 6 available place s,2nd girl can be se ate d in the re m aining 5 avaialble place s ,3rd girl can be se ate d in re m aining 4 available place s....e tc

ie 6 x 5 x 4 x 3 x 2 .

so total arrange m e nts boils down to : 2 * 5! *(6x 5x 4x 3x 2) Hope its cle ar now.

Ashwin

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by sharm ishtha Gupta - Saturday, 10 O ctobe r 2009, 10:03 PM Thanx ashwin.

U m ight think m e to be a dud but k indly e x plain how u got the 2 in 2*5!*(6*5*4*3*2) .

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by AsHwIn Drm z - Saturday, 10 O ctobe r 2009, 10:59 PM

he y..

dont think lik e dat..W e are all he re to le arn from e achothe r no one is pe rfe ct.So ne ve r unde re stim ate urse lf...C om ing to the que stion, se e the re will be 2 case s :1st case whe n u se at all the boys 1st in 5 place s the n arrange the girls .2nd case you can se at the girls 1st the n arrange the boys am ong the m .

he nce we tak e 2* Hope its cle ar now. Ashwin

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by Saravanar B - W e dne sday, 21 O ctobe r 2009, 01:24 PM

Hi TG Sir,

Thanks a lot for your article and I don't have any words to describe ur work sir...

I owe a lot... Great job sir... I have a great respect for you sir and long live TG family

Have a small doubt sir :

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Sol :

(approach 1) :

The first friend can stay in 10 ways, the second friend can stay in 10 ways, similarly the third

and fourth..

so the ans is

101010*10 = 10000

(approach 2) :

The first hotel can accomodate 4 friends or 3 friends or 2 friends or 1 friend or 0(zero). So nu

of ways the 1st hotel can accomodate is 1+2+3+4 = 10 ( i have added 1,2,3,4 as it is either - "OR" case)

Same goes for 2nd hotel, 3rd hotel and 4th hotel...

so the ans is

101010*10 = 10000

*******************************************************

Now have a look at question no 2

*******************************************************

2. In how many ways can u post 10 letters in 4 letterboxes ?

Sol :

( approach 1 )

The first letter can be posted in 4 ways, similarly the second,third....tenth.

So the ans is :

444...*4 = 4^10 = 1048576

(approach 2 ) :

The first letter box can get '1' or '2' or '3' ... '10' letters.So nu of ways

the 1st letter can accomodate is 1+2+3+4+..+10 = 55

Same goes for 2nd letter 3rd letter and 4th letter

so the ans should be 555555*55= 9150625

Doubt : I couldn't figure out what is wrong with my second approach.. Kindly explain me sir.. Also tell me

given a question when to apply which approach...

My prob is I never get satisfied with one approach sir.. I always try out different approach..

I am sure ur explanation would be helpful to many of the students like me...

Thanks

Saran

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by Total Gadha - W e dne sday, 21 O ctobe r 2009, 09:42 PM

Hi Saravanar,

If first letter box posts even 1 letter or 2 letters etc. the second letter box does not have 10 ways.

Total Gadha

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by sk ysk ie rs 44 - Monday, 26 O ctobe r 2009, 02:55 AM He llo,

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m not sure if this is the right se ction to post pe rm utation proble m , but anyways ple ase he lp m e out with the following In how m any ways can 10 soldie rs stand in 2 rows such that the re are 5 soldie rs in e ach row?

Is the answe r 7257600 or 7257600/2 ? Pune e t

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by Nabanshu Bhattacharje e - Tue sday, 27 O ctobe r 2009, 01:33 AM Hi TG

Thank s again,

W ante d to share the probabilistic approach to de arrange m e nt proble m with the TGite s.

http://m athforum .org/library/drm ath/vie w/56592.htm l

A que ry: If you go through the proof in the link I have give n, you can se e that the y are conside ring 4 out of 5 le tte rs place d in the corre ct e nve lope . How com e the re re sult m atche s with the one give n in this thre ad? I m e an i could not find whe re I am going wrong. Ple ase he lp!!!

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by Saravanar B - Friday, 30 O ctobe r 2009, 12:26 AM

Hi TG,

Thank s for your re ply...

Ye s i got it... the se cond le tte r box ne e d not have te n ways...

Is the re a way to solve this que stion in the letter box perspective (approach 2) ?? I trie d to solve but couldn't ge t the ans as we ll as the approach..

Saran

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by swe tha chinthare ddy - Monday, 2 Nove m be r 2009, 03:35 PM

6 Balls of diffe re nt colours are to be distribute d am ong 2 boys. W hat is the probability that e ach boy ge ts sam e num be r of balls? a) 5/16

b) 15/64 c) 1/2 d) 3/8

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by nishchai ne vre k ar - W e dne sday, 4 Nove m be r 2009, 05:58 PM Hi,

I would lik e to point out the that the solution to the 5 BO Y , 5 GIR L MO VIE m ovie proble m with no girls sitting toge the r is a little flawe d... It doe snt tak e in to conside ration this arrange m e nt

GBGBGBBGBG

n m any othe rs lik this....

so the soln to this has to be considered by keeping objects (or people for this problem) who are not supposed to be together as dynamic and others as static...

he re boys have to be static n girls dynam ic... as shown be low

_B_B_B_B_B_

so girls only have six positions to go into and also the y will ne ve r be toge the r as we lim iting the m by providing only one slot btw boys. He nce re fe r to the m as dynam ic as the y can tak e 6 positions.

so, 6P5 = 6x 120

while boys who can be toge the r are static cause the y can only tak e 5 positions. 5P5 = 120

so, total no ways = 120x 120x 6 = 14400x 6 Hope this was he lpful...

Also, Ple ase corre ct m e if I am wrong :D

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Hi Nishchal,

You forgot the condition mentioned in the question- every boy wants to sit with a girl.

Total Gadha

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by saurabh prabhude ssai - Friday, 6 Nove m be r 2009, 03:42 PM

in the condition m e ntione d by nischai e ve ry boy has at le ast one gal on his side !!

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by nishchai ne vre k ar - Friday, 6 Nove m be r 2009, 03:59 PM in the se type s of arrange m e nts

GB GB BG BG BG

Here every boy is sitting with one girl.... it the condition was

"every boy wants to sit in-between 2 girls" then i would be a diff case altogether.

If not then plz temme... y these cases are not to be considered...

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by vine e t jain - Tue sday, 17 Nove m be r 2009, 12:41 AM Hi TG!!

supe rb article

I have a que stion which has be e n ask e d by R ohit also.i am using sam e language as him "In how m any ways can 4 pe rsons be se ate d out of 5 boys and 3 girls in 4 diffe re nt se ats? W hile e x plaining you said it is not the sim ple form ula base d P&C proble m .

Though, I think that it could be solve d in ve ry sim ple te rm s without bre ak ing down the proble m into case s(the way you alre ady sugge ste d as the conce pt).

Two aspe cts to this proble m :

1. Se le cting 4 pe rsons from 8 pe rsons(5 boys+3 girls) which can be done in 8C 4 ways

2. Arranging the 4 pe rsons on 4 diffe re nt se ats for e ach se le ction which can be done in 4! ways So total num be r of ways = 8C 4 * 4! = 70*24 = 1680 (sam e as 8P4)"

ple ase corre ct m e if i am doing it wrong

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by SHAUNAK A - Tue sday, 24 Nove m be r 2009, 08:16 PM He llo TG sir,

That's C AT lik e article on Pe rm C om ...sim plicity e x e m plifie d... Thanx Sir...

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by Harish A - W e dne sday, 25 Nove m be r 2009, 09:32 PM

Hi TG,

This is My first post in this is forum .

It is gre at disappointm e nt I would lik e say, e ve n afte r going through the Le ssons m ultiple tim e s, I am unable to solve e ve n a single proble m . It is not be cause of the way you e x plain things, but it is m y ability.

I am sure with the se give n se t of sk ills, I am ill e quippe d to tack le C AT.

It m ay be of gre at he lp if you can le t m e whe re the solutions are pre se nt I can just re ad the m for purpose of le arning. Thank s,

Harish.A

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(37)

Hi TG

I love d this article ..thank s a lot for ur e fforts ..this he lpe d m e re colle ct m any of the conce pts le arnt e arlie r..

but i am not cle ar about the 'painting face s of a cube ' proble m ..two othe r guys Syd S and Anirban Bhar have also ask e d the sam e que stion that i am going to type now..

W he n we fix the color of one side of a cube ..we have 6 possibilitie s (as the re are six colors)..and in that case the opposite side will have 5 possibilitie s..for the re m aining we can use the circular pe rrm utation rule ..so the re m aining side s can be painte d in (4-1)! ways..

So shouldnt the num be r of ways in which the cube can be painte d, be 6x 5x 3! ? C an u ple ase clarify this?

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by gurte z singh - Thursday, 11 Fe bruary 2010, 05:56 PM nice work sir ...

i hve a prob...

~ the re r four se ctions of a pape r wid a m ax . m ark s f 45 fr e ach se ction & to qualify 1 have ge t m in. 90 m ark s the n in hw m any ways he cn qualify da e x am ??????????????

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by vijayshre e m e non - Monday, 5 April 2010, 10:05 PM Hi ,

C an any1 e x plain m e this

In the above post , TG sir cite d an e g

@ an election meeting 100 ppl address rally. Order followed PM then MP followed by MLA

Solution give n: Le t PM = P MP de note d as Q and MLA as R . 10 spe ak e rs adddre ss rally in 10! ways. this include s PQ R PR Q Q PR Q R P R PQ R Q P. Now Since ne e d only PQ R we divide 10! by 6.

I have a ge nuine prob he re . First of all its de cide d tat PQ R will be the orde r which m e ans we ne e d to k e e p PQ R fix e d and only count the re m aining 7 ppl arrange m e nt. Y cant it be so ??? I k now I wrong whe re am I m ak ing the m istak e ....

Im a re al dum b one to have such doubts ... but want to m ak e it cle ar .... Pls TG sir or Dagny Mam or who e ve r cam am swe r this q

Pls he lp Pls...

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by Aspire nd Achie ve - Monday, 10 May 2010, 02:27 PM Hi TG Sir,

can you ple ase e x plain the solution for proble m 3

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circular -linera query

by robin catch - Friday, 25 June 2010, 03:41 PM

DEAR TG..help me on this??

the number of ways that 4 girls and 5 boys can sit around

a table st no two girls sit together ?

is there a general formula??

what would be the solution for a linear table, ie 4 girls and

5 boys( no 2 girls sit together)??

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(case1:around a round table

case 2: in a linear arrangement)???

thnx in advance, highly appreciate for quick reply

imlovinmath

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A nswers

by R am it Goyal - Monday, 28 June 2010, 11:04 PM 1 a 2 b 3 d 4 c 5 c 6 a 7 d 8 c 9 c 10 b 11 b 12 c 13 14 a 15 16 17 b 18 d 19 a 20 c

Ple ase re vie w and notify m e for unatte m pte d que stions and wrong answe rs

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by sonali pawar - Thursday, 8 July 2010, 03:12 PM Hi TG,

As usual be autiful article !!!! Just one doubt...

In the part of Distribution in the proble m of 5 balls and 3 box e s case IV i.e whe n all balls and box e s are dissim ilar, for distribution 113 and 122 u say the no of distributions are 3!. Shouldn't it be 3!/2! = 3 as the re are 2 groups with sam e num be r of things?

Ple ase corre ct m e if m wrong!

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by TG Te am - Thursday, 8 July 2010, 05:32 PM

Hi Sonali

Though in 113 and 122 distribution there are two groups with same number of things but still they are different

groups because of different balls. That's why it is 3! in place of 3!/2!

Hope it is clear.

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by Nam an Mirchandani - Tue sday, 20 July 2010, 04:29 PM

Sir,

(39)

In how many ways can 5 similar balls be

distributed in 3 different boxes?

Ans: (by method given here) 21

But there is one basic question in you

article (like in how many can one post 10 letters

in 4 letter boxes: 4 4 4...10 times :

4^10....)...

So if we apply the above method considering

similar letters and different letterboxes

...answer will come : 13C3 ?

Have we considered all letters dissimilar to get

4^10? (can it be generalised where nothing is

mentioned about similarity or dissimilarity ? )

Pls help ??????????? ....

Naman

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by Siddharth Khanna - Saturday, 7 August 2010, 09:36 PM Hi TG Sir ,

(

in the se type s of arrange m e nts

GB GB BG BG BG

Here every boy is sitting with one girl.... it the condition was

"every boy wants to sit in-between 2 girls" then i would be a diff case altogether.

If not then plz temme... y these cases are not to be considered...

)

Eve n I have the sam e doubt as m e ntione d by nishchai . C ant we conside r this arrange m e nt .

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by ravi baranwal - Sunday, 8 August 2010, 03:32 AM agre e wid siddharth n nischai.

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by k e sav singh - W e dne sday, 11 August 2010, 11:01 PM

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can anybody he lp m e in be low que stion?

how m any diffe re nt sum s can be form e d with the following coins

5 rupe e ,1 rupe e ,50 paise , 25 paise , 10 paise , 3 paisa , 2 paisa & 1 paise ? plz e le borate

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by R avi Mathur - Tue sday, 17 August 2010, 09:16 PM

The re are 10 ste ps in a staircase and a pe rson has to tak e those ste ps. At e ve ry ste p the pe rson has got a choice of tak ing 1 ste p or 2 ste ps or 3 ste ps.The num be r of ways in whch pe rson can tak e those ste ps is??

C an anyone ple ase he lp m e with this staircase proble m .

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by ve rsha jhunjhunwala - Sunday, 29 August 2010, 12:05 AM

O ut of 21 tick e ts m ark e d with num be rs from 1 to 21 ,thre e are drawn at random , find the probability that the thre e num be rs on the m are in A.P

can u ple ase e x plain the com ple te m e thod?

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by NITIN NIGAM - Tue sday, 31 August 2010, 07:57 AM Hi Ve rsha...m y approach is as follows:

from 1 to 21 find the no. of AP's that u can ge t and the re will be case s accordingly:

1. AP with d = 1 , we have 21 nos. Am ong the se 21 nos. we can choose 3 nos. which are in AP as 1,2,3 or 2,3,4 or 3,4,5 and so on...he nce total no. of case s will be 19 he re with last case e nding at 19,20,21

2. AP with d = 2, we have two se rie s he re a) 1,3,5,7...21 = 11 nos.

he nce nos. can be chose n as 1,3,5 or 3,5,7 and so on ..so he re TO TAL C ASES = 9 b) 2,4,6,8...20 = 10 nos.

he nce nos. can be chose n as 2,4,6 or 4,6,8 and so on..so he re TO TAL C ASES = 8 3. AP with d = 3, we have the following se rie s

a) 1,4,7,10,13,16,19 -> total case s = 5 b) 2,5,8,11,14,17,20 -> total case s = 5 c) 3,6,9,12,15,18,21 -> total case s = 5 4. AP with d = 4, we have the following se rie s a) 1,5,9,13,17,21 -> total case s = 4 b) 2,6,10,14,18 -> total case s = 3 c) 3,7,11,15,19 -> total case s = 3 d) 4,8,12,16,20 -> total case s = 3

5. AP with d = 5, we have the following se rie s a) 1,6,11,16,21 -> Total case s = 3

b) 2,7,12,17 -> Total case s = 2 c) 3,8,13,18 -> Total case s = 2 d) 4,9,14,19 -> Total case s = 2 e ) 5,10,15,20 -> Total case s = 2

6. AP with d = 6, we have the following se rie s a) 1,7,13,19 -> Total case s = 2 b) 2,8,14,20 -> Total case s = 2 c) 3,9,15,21 -> Total case s = 2 d) 4,10,16 -> Total case s = 1 e ) 5,11,17 -> Total case s = 1 f) 6,12,18 -> Total case s = 1

7. AP with d = 7, we have the following se rie s a) 1,8,15 -> Total case s = 1 b) 2,9,16 -> Total case s = 1 c) 3,10,17 -> Total case s = 1 d) 4,11,18 -> Total case s = 1 e ) 5,12,19 -> Total case s = 1 f) 6,13,20 -> Total case s = 1 g) 7,14,21 -> Total case s = 1

8. AP with d = 8, we have the following se rie s a) 1,9,17 -> Total case s = 1

b) 2,10,18 -> Total case s = 1 c) 3,11,19 -> Total case s = 1 d) 4,12,20 -> Total case s = 1 e ) 5,13,21 -> Total case s = 1

9. AP with d = 9, we have the following se rie s a) 1,10,19 -> Total case s = 1