- T.N.Kaul
- J.R.D. Tata
- Khushwant Singh
- Nani Palkhivala

OA : D

2. Who is the author of the book ‘Nineteen Eighty Four’?

- Thomas Hardy
- Emile Zola
- George Orwell
- Walter Scott

OA : C

3. Which of the following is NOT written by Munshi Premchand?

- Gaban
- Godan
- Guide
- Manasorovar

OA : C

4. Who is the author of the book ‘Forbidden Verses’?

- Salman RushDie
- Abu Nuwas
- Ms. Taslima Nasrin
- D.H. Lawrence

OA : B

5. The lead character in the film ‘The Bandit Queen’ has been played by

- Rupa Ganguly
- Seema Biswas
- Pratiba Sinha
- Shabama Azmi

OA : B

6. Which of the following books has been written by Vikram Seth?

- My God Died Young
- Islamic Bomb
- Look Back in Anger
- A Suitable Boy

OA : D

7. Who wrote the line: ‘ A thing of beauty is a joy forever’?

- John Keats
- Robert Browing
- P.B.Shelley
- William Wordsworth

OA : A

8. Who has written the book ‘My Frozen Turbulence in Kashmir’?

- Anees Jung
- Jagmohan
- M.J.Akbar
- Nissim Ezekiel

OA : B

9. The book ‘To Live or Not Live’ was written by

- V.S. Naipaul
- Alexandra Dumas
- George Elliot
- Nirad C. Chaudhuri

OA : D

10. In India the first television programme was broadcasted in

- 1959
- 1965
- 1976
- 1957

OA : A

11. ‘A Voice for Freedom’ is a book written by

- Corazon Aquino
- Nayantara Sahgal
- Aung San Suu Kyi
- Benazir Bhutto

OA : B

12. Who is the father of English Poetry?

- Milton
- Wordsworth
- Chaucer
- Charles Dickens

OA : C

13. The popular TV serial ‘Neem ka Ped’ was written by

- Rahi Masoom Raza
- Sharad Joshi
- Ashapurna Devi
- Harivansh Rai Bachachan

OA : A

14. Who wrote ‘War and Peace’?

- Leo Tolstoy
- Mahatma Gandhi
- Charles Dickens
- Kipling

OA : A

15. Which of the book was written by Graham Greene?

- O’ Jerusalem
- The Power and Glory
- Crisis in India
- Of Human Bondage

OA : B

16. George Bernard Shaw, the great dramatist, was

- a Welsh
- an Irishman
- a Soctsman
- an Englishman

OA : B

17. ‘Mirchchhakatika’ was written by

- Vikramaditya
- Shudraka
- Kalhana
- Banabhatta

OA : B

18. The famous book ‘Anandmath’ was authored by

- Sarojini Naidu
- Bankim Chandra Chottapadhya
- Sri Aurobindo
- Rabindrnath Tagore

OA : B

19. Who is the author of “Divine Comedy”?

- Goethe
- Milton
- Dante
- Shakespeare

OA : C

20. Who is the author of the book ‘Chronicle of a Corpse Bearer’?

- Cyrus Mistry
- KR Meera
- Jaswinder Singh
- Madhu Acharya

OA : A

]]>You have 3 sections

1. Verbal Ability : This section includes Reading comprehensions, Para Jumbles and Para Completion ( This is according to the recent trend ).

Number of questions : 34 Time : 60 minutes.

2. DILR : This section includes almost 4 DI and 4 LR sets. 1 set has about 3 – 4 questions.

The syllabus is vast for this one.

**LR – DI TOPICS**

For OMETs

LOGICAL REASONING :

● CUBES 2 – 3

● SEATING ARRANGEMENTS – 2

● COMPLEX ARRANGEMENTS – 2

● CONDITIONS & GROUPING – 2

● CODING & DECODING – 2

● SEQUNECES AND SERIES ( NUMBERS AND LETTERS ) – 2

● BINARY LOGIC – 2

● CALENDARS & CLOCK – 2

● INPUT OUTPUT – 2 – 3

● DIRECTION SENSE TEST – 2

● COMPARISON OF RANKS – 1

● BLOOD RELATIONS – 1

● MISCELLANEOUS CLASSES

DATA INTERPRETATION :

● Tables – 2

● Bar Charts – 2

● X-Y Charts Involving Line Series – 2

● Pie Charts – 2

● Cases – 2

For CAT / XAT/ IIFT :

higher level of the basics covered in OMETs + :

● TABULAR QUESTIONS ( has the maximum weightage for CAT )

● CHESSBOARD

● GRID

● CUBES & CUBOIDS

● SET THEORY

● GAMES & TOURNAMENTS

● MAXIMA & MINIMA

● CRITICAL PATH

● COIN GAME BASED

● SCHEDULING

● BINARY LOGIC

● SPRING BALANCE

● MATRIX

● CRYPTO – ARITHMETIC SERIES

● MISCELLANEOUS

^This is the list of some. I have tried to cover the most of it but again you know how LRDI has been varying and there may be some additional topics to this list.

Number of questions : 32 Time : 60 minutes

3. Quantitative Aptitude :

Topics :

1) Averages

2) Mixtures & Alligations

3) Percentages

4) Profitl Loss

5) Intrest

6) Ratio and Proportion

7) Time n Work

8) TSD

9) PnC

10) Probability

11) Logarithms

12) Graphs n Functions

13) Theory of Equations

14) Progressions

15) Inequalities

16) Number System

17) Set theory

18) Geometry

19) Mensuration

20) Miscellanous

● Geometry Mensuration subtopics

1) Line and Angles

2) Triangles

3) Polygons

4) Circles

5) Areas

6) 2D Areas

7) Circles connected Mensuration

8) Volumes and Surface Areas

● Number System Subtopics :

1) Unit Digit

2) Last two Digits

3) HCF LCM

4) Factors

5) No. Of Zeros

6) Remainders

7) Base system

8) Integral Solutions

9) Miscellanous

● Theory of Equations subtopics

1) Basics of Algebra

2) Polynomials

3) Linear Equations

4) Quadratic Equations

5) Solution and Location of roots.

This is the total syllabus.

How to prepare :

GENERAL FOR ALL THREE SECTIONS : Follow these 4 steps:

1st prepare basics from the any institute’s material. Have a fair idea on how to go about the question and how to approach them.

2nd : The more you pratuce the more you will score and improvise. Practice higher level questions. Don’t go too deep into it some people do olympiad level questions which arr not necessary CAT always ask basic questions.

3rd : Now start solving previous year papers Do these religiously.

4th: Most important – MOCKS . Take as many mocks as possible and analyse them. Remember that Analysing is really important.

See where you are making mistakes and make sure you dont repeat them again.

Apart from this start reading The hindu editorial daily and Norman lewis book for vocab.

You can also join our coaching GENIUS TUTORIALS : www.geniustutorials.guru to attend regular classes and have a proper CAT level study plan.

]]>**Make Weekly Targets And Celebrate Your Achievements :**The best way to motivate yourself is to set small targets and keep a track of your activities. Completion of one task itself acts as fuel for next week.**Wake up Early Morning :**“ The early morning has gold in its mouth,” said Benjamin Franklin. When a person gets up early, he is more energetic and takes lesser time to do a task that would take more time otherwise. It reduces your stress level and you can start your day at an optimistic note ( without any rush ). Getting up early instils you with positivity.**Keep Participating In All Competitions :**Its okay if you are preparing for CAT but you should keep taking other exams also like Banking Sector / SSC etc. to keep your competitive spirits alive.Results may or may not be favourable but facing real exam develops your skill to handle pressure which is something you cant get in your mock test series. Additionaly if you clear you can get a backup option also.**Balance all Sections & Avoid to practice LOD 10 in QA :**CAT is an exam that tests your smartness and iq and i experience that many students keep themselves busy in some Olmpiad Level Problems. Please avoid such blunders. Instead its better if you give equal time to every subject especially VA section as it demands regular reading practice at your end.**Workout & Eat Healthy :**When you study, your brain consumes glucose, so take a five-minute break every hour to let your body produce more fuel for your studying. Eating a healthy snack is very beneficial and can make a significant difference (almonds, fruit, and yogurt are good choice. Avoid taking oily/junk food and should go for workout daily. Exercise will increase your energy level and reduce the effects of stress.**Believe in Yourself :**Believe that you can do it regardless of what anyone says or where you are in life. Think about exactly what your life would look like if you had already achieved your dream. It is important to always hold a positive expectation that what you want is going to happen. It is simply a choice and a discipline of the mind.**Read Success Stories :**History is full of stories of people who also came into hard times, but they made a very different set of choices in their lives. Rather than losing hope and giving up, they chose to pick themselves up, brush themselves off, and continue moving with the hope toward achieving their dreams. These stories are your fuel for this season.**Most Important :**CAT is just an exam or its just a way out of thousands ways to be successful in career. Give your best shot and leave no stone unturned during your preprations but never make it a matter of Ego/Insult. If you are good you ll definitely make another way to . “We must accept finite disappointment, but we must never lose infinite hope.” –Martin Luther King Jr.

just basic knowledge of mathematics or from non – mathematics backgrounds are usually afraid

of CAT or other management entrance competitions. But as far as QA is concerned you need

not to go for much rocket science study to excel in this domain.

Before discussing preparation strategy, we must have knowledge of certain eligibilities and

parameters for getting into our dream B – school.

ELIGIBILITIES

Candidates should have done graduation in any stream.

Minimum marks in graduation are 50% and 45% for reserved category.

Final year students can also apply.

RESERVATIONS

For SC candidates – 15%

For ST candidates – 7.5%

For OBC candidates – 27%

For PWD candidates – 3%

CAT preparation strategy is different for different candidates.

Candidates can be

Working Professional

Final year students

Re appearing for CAT

In a do or die situation with last attempt for CAT

Apart from CAT score other parameters are there for selection like

Your score in X , XII and graduation levels

Your stream in graduation

Your internships or NGO’s joined if any

Your Work ex/ Gender/ Gap Year etc

Main question students usually ask me is

Whether to quit job?

See let us point out some advantages /disadvantages of quitting job

ADVANTAGES

You are free from whole day distractions that you face during job.

You get a lot of time for preparation.

During job, when you start studying you are already exhausted after full day work in job.

But now you are energetic to start your preparation.

DISADVANTAGES

Any competition and especially CAT is unpredictable as even after 99+ percentile it’s not

sure that you get your dream B – school or not. So, quitting job will be a risky situation in

that case.

Preparing for CAT just after quitting will be a Gap year between your studies and job, and

it would be hard to justify during interviews.

Last but not the least, you are going to be a Manager and preparing for CAT along with

your Job itself is an example of good management which can have positive effect in later

stages.

DO’S AND DON’TS

If starting from zero, give 15 – 20 days to NCERT books questions for basics.

Never ever go for direct shortcuts. Even if you are using any shortcuts you must know the

logic/proof behind that shortcut.

Try to understand proofs of all geometry theorems because many conceptual questions

are based on it.

Never confuse your preparation with multiple sources or materials. Just follow any one

content and solve it religiously.

Each student has a specific weal area and a strong area. Try to give maximum time to your

weak topics in initial stage of preparation.

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What does the course include?

● Live online classes with the recordings

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● Effective doubt sessions

Now, the important question, why should you choose us? or, how are we different?

1. MANTHAN covers special zero hour sessions for beginners and non-maths background students.

2. Individual attention – When we say this, we not only promise you to solve your individual doubts but also to guide you at every phase of your preparation.

3. On regular intervals, we’ll keep conducting some special sessions on shortcuts and handy tricks for the exam.

4. BEST PART OF MANTHAN’19 – The full course includes GK as well. We all know the importance of GK in OMETs, but we always begin late. This year, stay updated from Day 1 of your preparation.

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ABC, ACB, BCA, BAC, CBA, CAB

Now that was easy to figure out as less number of letters are there. But if we have to find that how many 6 letter names can be made out of English alphabets with certain conditions that would be time consuming. Here comes the need and application of this chapter. Permutations and combinations is a very interesting chapter and gives us food for thought

**Fundamental principle of Addition**

If there are two jobs such that they can be performed independently in m and n ways, then either of the two jobs can be performed in (m + n) ways.

**Fundamental principle of Multiplication**

If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m x n ways.

Example 1 : How many words of three distinct letters of English alphabets are there?

Answer : 26 * 25 * 24

Example 2 : How many numbers lying between 5000 and 6000 (both inclusive), can be formed if,

(i) Repetition of digits is not allowed.

(ii) Repetition of digits is allowed.

If we are to form the numbers without repeating any digit then we can’t take 5000 and 6000 as 0 is repeating. Now if number is between 5000 and 6000, we can say that thousand’s place can take only one digit that is 5. For hundred’s place we are left with only 9 digits, for ten’s place only 8 digits and for unit’s place only 7 digits.

So the total number of numbers between 5000 and 6000 in which no digit is repeating will be 9 x 8 x 7 = 504

Now if the repetition of digits is allowed, we can include all the numbers right from 5000 up to and with 6000. So the total number of numbers will be 1001

**Number of ways to arrange r thing out of n, nPr = n!/(n-r)!**

For example : Number of ways in which 7 students can be seated on 10 chairs in 10P7.

How many words can be formed using all letters of word GENIUS?

Here since we are using 6 letters. So 6 letters can be arranged in 6 place in 6P6 = 6! Ways

How many words can be formed using all letters of the word ORIENTAL

OA: 8!

In how many ways can letters of word GENIUS be arranged so that the word must start with N?

Here since first place is fixed as N. So remaining 5 letters can be arranged in 5 places in 5! Ways

How many of the words that can be formed using all letters of the word ORIENTAL will start with N?

OA: 7!

How many of the words that can be formed using all letters of the word ORIENTAL will start with N and end with L?

OA: 6!

In how many of the words that can be formed using all letters of the word ORIENTAL will all vowels come together?

Now here as we want those words where all vowels come together. so put all vowels in a box and consider it as a single letter. now we have 5 letters and these 5 letters can be arranged in 5! ways and also 4 vowels in that box can also be arranged in 4! ways so answer should be 5! * 4!

In how many of the words that can be formed using all letters of the word ORIENTAL all vowels do not come together?

Total words — all vowels together.

so 8! — 5! * 4!

In how many ways can letters of word GENIUS be arranged so that no two vowels come together?

Here XGXNXSX. So 3 vowels can be arranged in 4 places in 4p3 ways and consonants themselves can be arranged in 3! ways. Hence 4p3 * 3!

In how many of the words that can be formed using all letters of the word ORIENTAL, no two vowels come together?

OA: 5p4 * 4!

In how many of the words that can be formed using all letters of the word ORIENTAL vowels occupy even places?

Now 4 vowels can be arranged at 4 even places in 4! ways and same way 4 consonants in 4! ways at odd places. so answer should be 4! * 4!

In how many words that can be formed using all letters of the word ORIENTAL vowels and constants come alternatively?

As we did for even, same is for odd places. so total should be 2 × 4! × 4!

Now suppose you have to choose 3 letters from 5 letters then selecting A, B, C and C, B, A are different cases?

Obviously no. here comes the concept of combinations. whenever we need to select few things we use combinations and when we need to arrange we use permutations.

**Difference between permutation and combination :**

- In a combination only selection is made where as in a permutation not only selection is made but also an arrangement in a definite order is considered.
- In a combination, the ordering of the selected objects is immaterial where as in a permutation the ordering is essential. For example A,B and B,A are same as combination but different as permutations
- Practically to find the permutations of n different items, taken r at a time, we first select r items from n items and then arrange them. So, usually the number of permutation exceed the number of combinations.
- Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, BAC, CBA & CAB correspond to the same combination ABC.

In how many ways can a cricket eleven be chosen from 15 players if

i) a particular player is always chosen

ii) a particular player is never chosen

OA: 14C10, 14C11

**The number of permutations of n things of which p1 are alike of one kind; p2 are alike of second kind; p3 are alike of third kind … pr are alike of rth kind such that p1 + p2 … + pr = n is n!/p1!p2!…pr!**

Find the number of ways in which 6 boys and 6 girls can be seated in a row so that

a) All the girls sit together and all the boys sit together

b) All the girls are never together

OA: a) 2 * 6!6! b) 12! — 7!6!

The number of ways in which the letters of the word ARRANGE be arranged so that

a) Two R’s are never together

b) Two A’s are together but not two R’s

c) Neither two A’s nor two R’s are together.

Solutions:

a) [RR], A,A,N,G,E = > 6!/2!

and total = > 7!/2!2!

so required answer = 7!/2!2! — 6!/2! = 900.

b) Both A’s together → [AA],R,R,N,G,E = 6!/2! = 360

Both A’s and both R’s together → [AA],[RR],N,G,E

= 5! = 120

required answer = 360–120 = 240.

c) Neither two A’s nor two R’s together → 900–240 = 660

How many chords can be drawn through 21 points on a circle ?

A chord is obtained by joining any two points on a circle. Therefore total number of chords drawn through 21 points is same as number of ways of selecting 2 points out of 21 points. Hence number of chords can be drawn in 21C2 = 210 ways.

There are 10 points in a plane, no three of which are in the same straight line, except 4 point which are colinear. Find the

a) Number of straight lines that can be formed

b) Number of triangles that can be formed using these

Solutions:

a) Number of straight lines that can be formed if all 10 are non collinear = 10C2 = 45

But 4 of them are collinear. So number of straight lines that can be formed using 4 points = 4C2 = 6

Now, 4 points when joined pairwise give only one line.

So, total lines = 45–6 + 1 = 40

b) Number of triangles that can be formed by joining 10 points = 10C3 = 120

Number of triangles formed by joining the 4 points, taken 3 at a time = 4C3 = 4

So total number of triangles = 120–4 = 116

N is an integer such that 1⁰⁷ ≤ N ≤ 1⁰⁸ and sum of digits of N is 3. How many such numbers are possible?

N is an 8 digit number and its left most digit can be 1, 2 or 3.

- When left most digit is 1: Other two digits can be two 1’s and 5 0’s or one 2 and six 0’s.

Number of ways possible for (Two 1s and five 0s): 7!/5!2! = 21

Number of ways possible for (one 2 and 6 0’s): 7!/6! = 7

Therefore, 21 + 7 = 28 cases possible. - When leftmost digit is 2:

One of the other digit can be 1 so one 1 and six 0’s

Number of ways possible = 7!/1!6! = 7

7 cases possible here. - When leftmost digit is 3:

All other digits have to be 0

So only 1 case possible

So total cases = 28 + 7 + 1 = 36

There are six boxes numbered 1,2,3… 6. Each box is to be filled up either with a red or a green ball in such a way that atleast one box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

a) 5

b) 21

c) 33

d) 60

Number of ways of filling 1 green ball: 6

[(1),(2),(3),(4),(5),(6)]

Number of ways of filling 2 green balls: 5

[(1,2),(2,3),(3,4),(4,5),(5,6)]

Number of ways of filling 3 green balls: 4

[(1,2,3) (2,3,4) (3,4,5) (4,5,6)]

Number of ways of filling 4 green balls: 3

[(1,2,3,4) (2,3,4,5) (3,4,5,6)]

Number of ways of filling 5 green balls: 2

[(1,2,3,4,5) (2,3,4,5,6)]

Number of ways of filling 6 green balls : 1

[(1,2,3,4,5,6)]

Hence required number of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21

How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS?

Total Letters : 11 (MM,AA,TT,H,E,I,C,S)

- All distinct : 8C4 x 4! = 1680 ( 8C4 because we have to select 4 out of 8 distinct letters M,A,T,H,E,I,C,S and 4! For the number of ways to arrange these)
- Two distinct and Two alike : 3C1 7C2 x 4!/2! = 756 ( 3C1 to select any 1 pair from (MM,AA,TT) 7C2 to select any two letters from the remaining 7 distinct letters and 4!/2! To arrange them)
- Two alike of one kind and two alike of another kind : 3C2 x 4!/2!2! = 18

So total number of ways = 1680 + 756 + 18 = 2454.

**Number of ways to select r objects out of n distinct objects in nCr**

In how many ways can we select 2 balls from a bag of 5 balls ( 1 Red, 1 Green, 1 Blue, 1 Yellow, 1 Orange) ?

Answer : 5C2

But, in how many ways can we select 2 balls from a bag of 5 red balls ?

since all balls are red in color so answer should be 1

Number of ways of selecting r objects out of n identical objects is 1

In how many ways can 5 letters be posted in 4 letter boxes ?

In this type of questions, one thing is repeatable and the other one is non-repeatable.**So our answer is always : repeatable ^ non repeatable**

So here the letter boxes are repeatable and the letters are not so our answer is ⁴⁵

**Number of ways to arrange r things out of n when repetition is not allowed are : nPrNumber of ways to arrange r things out of n when repetition is allowed are : n^r**

Find the number of 3-digit numbers than can be formed using the digits from 1–7 when repetition of digits not allowed.

Answer : Here repetition is not allowed so our answer will be of the form nPr.

N = 7 and r = 3

So our answer is 7P3

Find the number of 3 digit numbers that can be formed using the digits from 1–7 when repetition of digits is allowed.

Answer : Here repetition is allowed so our answer will be of the form n^r

N = 7 and r = 3

So, our answer is ⁷³

**Number of ways to arrange n items in a circle is (n-1)!**

A circle has no end points. If there are n items, let us say we place 1 of them at a fixed position. Now the end points are defined. So the number of ways in which (n-1) items can be arranged are (n-1)!

**Number of ways to arrange n items in a garland is (n-1)!/2**

A garland has no end points. If there are n items let us say we place 1 of them at a fixed position. Now the end points are defined so the number of ways in which (n-1) items can be arranged are (n-1)! But this is to be divided by 2 as clockwise and anti-clockwise arrangements are same. So total ways in which they can be arranged are (n-1)!/2

Five boys and five girls sit alternatively around a round table. In how many ways can this be done?

Answer : Five boys can be arranged in a circle in 4! Ways

After that girls can be arranged in the 5 gaps between boys in 5! Ways

Hence total number of ways = 4! * 5! = 2880

8 chairs are arranged in a circular arrangement numbered 1 to 8. In how many ways can 4 boys and 4 girls be seated on them such that all girls must be seated on even numbered chair?

Answer : If chairs are numbered then the arrangement is no longer considered as circular as here end point is defined. So consider it as a linear arrangement. Thus answer should be 4! * 4!

A round table conference is to be held among 20 delegates belonging from 20 different countries. In how many ways can they be seated if two particular delegates are

a) Always to sit together

b) Never to sit together

a) Let the two particular delegates who wish to sit together to be treated as one delegate. So we have 19 delegates who can be arranged on a round table in (19–1)! = 18! Ways.

Also the two particular candidates can be arranged among themselves in 2! = 2 ways.

So total number of arrangements = 2 x 18!

The number of arrangements of 20 delegates on a round table is (20–1)! = 19!

b) Number of arrangements where 2 particular candidates never sit together = (total arrangements — When two are always together)

= 19! — 2 x 18!

= 17 x 18!

12 persons are to be arranged around two round tables such that 1 table can accommodate 7 persons and another one can accommodate 5 persons. The number of ways in which these 12 persons can be arranged is

a) 12C5 x 6! X 4!

b) 6! X 4!

c) 12C5 x 6!

d) None of these

7 persons can be selected for the final table in 12C7 ways. Now these 7 persons can be arranged in 6! Ways.

The remaining 5 persons can be arranged on the second table in 4! Ways.

Hence the total number of ways = 12C5 * 6! * 4!

12 persons are to be arranged around two round tables such that 1 table can accommodate 7 persons and another one can accommodate 5 persons. The number of ways of arrangements possible if two particular persons A and B do not want to be on the same table is

a) 10C4 x 6! X 4!

b) 10C6 x 6! X 4! X 2

c) 11C6 X 6! X 4!

d) None of these

A can sit on first table and B on the second one or A on the second table and B on the first table.

If A is on the first table, then remaining six for the first table can be selected in 10C6 ways.

Now these 7 persons can be arranged in 6! Ways.

Remaining 5 persons can be arranged on the other table in 4! Ways.

Hence total number of ways are 10C6 * 6! * 4! * 2

12 persons are to be arranged around two round tables such that 1 table can accommodate 7 persons and other one can accommodate 5 persons. The number of arrangements possible if two particular persons A and B want to be together and consecutive is

a) 10C7 x 6! X 2! X 3! + 10C5 x 4! X 5! X 2!

b) 10C5 x 6! X 3! + 10C7 x 4! X 5!

c) 10C7 x 6! X 2! + 10C5 x 5! X 2!

d) None of these

If A, B are on the first table then remaining 5 can be selected in 10C5 ways.

Now 7 persons including A and B can be arranged on the first table in which A and B are together in 2! 5! Ways. Remaining 5 can be arranged on the second table in 4! Ways.

Total number of ways is 10C5 * 4! * 5! * 2!

If A, B are on second table then the remaining three can be selected in 10C3 ways.

Now 5 persons including A and B can be arranged on the second table in which A and B are together in 2! 3! Ways.

Remaining 7 can be arranged on the first table in 6! Ways

Hence the number of ways for the first table is 10C7 * 6! * 2! * 3!

Answer = 10C7 * 6! * 2! * 3! + 10C5 * 4! * 5! * 2!

**Number of Squares and Rectangles in a Grid**

Consider a 2 x 2 grid:

Number of rows = Number of columns = 2

Number of squares = 4 squares of side (1 x 1) + 1 square of side 2×2 = 4 + 1 = ²² + ¹² = 5

Number of vertical lines = Number of horizontal lines = 3

So, number of rectangles = 3C2 x 3C2 = 3 x 3 = 9

Consider a 3 x 3 grid

Number of rows = Number of columns = 3

Number of squares = 9 squares of side (1×1) + 4 squares of side (2×2) + 1 square of side (3×3) = ³² + ²² + ¹² = 14

Number of vertical lines = Number of horizontal lines = 4

So number of rectangles = 4C2 x 4C2 = 6 x 6 = 36

**General formula for the number of squares for a n x n grid = ¹² + ²² + … + n²General formula for the number of rectangles for a n x n grid = (n+1)C2 x (n+1)C2 = ¹³ + ²³ + …. + n³**

Find the number of squares & rectangles in a chess board:

In a chess board, number of rows = number of columns = 8

So number of squares = ¹² + ²² + … + ⁸² = 8(8+1)(16+1)/6 = 204

Number of vertical lines = Number of horizontal lines = 9

Number of rectangles = 9C2 x 9C2 = 36 x 36 = 1296

**Derangement**

**If n distinct objects are arranged in a row, then the number of ways in which they can be deranged so that none of them occupies its original place is:D(n) = n!{1–1/1! + 1/2! — 1/3! + 1/4! — …. + (-1)^n 1/n!}**

**If r (0 ≤ r ≤ n) objects occupy the places assigned to them i.e none of the remaining (n-r) objects occupy their original place, then the number of possible ways:D(n-r) = nCr D(n-r)= nCr (n-r)! [ 1–1/1! + 1/2! — 1/3! + … + (-1)^(n-r)/(n-r)!]**

**Number of ways in which n items can be deranged: n! { 1–1/1! + 1/2! — 1/3! + 1/4! — … + (-1)^n 1/n!}**

Number of ways in which 1 item can be deranged:

1! {1–1/1!} = 0 (Obviously as only 1 item cannot be deranged)

Number of ways in which 2 items can be deranged: 2! { 1–1/1! + 1/2!} = 1

Number of ways in which 3 items can be deranged: 3! { 1–1/1! + 1/2! — 1/3!} = 2

Number of ways in which 4 items can be deranged: 4! { 1- 1/1! + 1/2! — 1/3! + 1/4!} = 9

Number of ways in which 5 items can be deranged: 5! { 1–1/1! + 1/2! — 1/3! + 1/4! — 1/5!} = 44

The number of ways in which 5 letters can be placed in 10 marked envelopes, so that no letter is in the right envelope is

Answer : D(5) = 44

Seven people leave their bags outside a temple and picked one bag at random after returning from worship. In how many ways can at least 1 and at most 3 of them get their correct bags?

Answer : Number of ways in which 1 get his own bag

= 7C1 6! [ 1 – 1/1! + 1/2! – 1/3! + … + 1/6!]

= 7C1 x 265

Number of ways in which 2 get their own bag

= 7C2 5![1 – 1/1! + 1/2! — 1/3! + 1/4! — 1/5!]

= 7C2 x 44

Number of ways in which 3 get their own bag

= 7C3 4! [ 1–1/1! + 1/2! — 1/3! + 1/4!]

= 7C3 x 9

Required number of ways = 7C1 x 265 + 7C2 x 44 + 7C3 x 9

Six cards and six envelopes are numbered 1,2,3,4,5,6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope number 2. Then the number of ways it can be done is

Answer : If 2 goes in 1, then it is derangement of 4 things which can be done in 4!(1–1/1!+1/2!-1/3!+1/4!)= 9 ways.

If 2 doesn’t go in 1 it is derangement of 5 things which can be done in 5!(1–1/1!+1/2!-1/3!+1/4!-1/5!) = 44 ways

Hence total 53 ways are there.