Gate Maths

SRI MANAKULA VIANAYAGAR ENGINEERING COLLEGE

1. From a group of 7 men and 6 women, five persons are to be selected to form a committee

so that at least 3 men are there on the committee. In how many ways can it be done?

A. 564 B. 645

C. 735 D. 756

Answer: Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)

=7 x 6 x 5

x6 x 5

+ (7C3 x 6C1) + (7C2)3 x 2 x 1 2 x 1

= 525 +7 x 6 x 5

x 6 +7 x 6

3 x 2 x 1 2 x 1

= (525 + 210 + 21)

= 756.

2. In how many different ways can the letters of the word 'LEADING' be arranged in such a

way that the vowels always come together?

A. 360 B. 480

C. 720 D. 5040

Answer: Option C

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

REAVATHY.N, ASST PROF, MATHS DEPT


3.

3. In how many different ways can the letters of the word 'CORPORATION' be arranged

so that the vowels always come together?

A. 810 B. 1440

C. 2880 D. 50400

Answer: Option D

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7!

= 2520.2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in5!

= 20 ways.3!


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4. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be

formed?

A. 210 B. 1050

C. 25200 D. 21400

Answer: Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)

=7 x 6 x 5

x4 x 3

3 x 2 x 1 2 x 1

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging = 5!



5 letters among themselves

= 5 x 4 x 3 x 2 x 1

= 120.


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5. In how many ways can the letters of the word 'LEADER' be arranged?

A. 72 B. 144

C. 360 D. 720

Answer: Option C

Explanation:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways =

6!

= 360.(1!)(2!)(1!)(1!)

(1!

6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different

ways can they be selected such that at least one boy should be there?

A. 159 B. 194

C. 205 D. 209

Answer: Option D

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number

of ways= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)

= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)

= (6 x 4) +

6 x

5x

4 x 3+

6 x 5 x

4x 4 +

6 x 5

2 x

1

2 x 1 3 x 2 x

1

2 x 1



= (24 + 90 + 80 + 15)

= 209.

7. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are

divisible by 5 and none of the digits is repeated?

A. 5 B. 10

C. 15 D. 20

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there

is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there

are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4

ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

8. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8

men and 10 women?

A. 266 B. 5040

C. 11760 D. 86400

Answer: Option C

Explanation:

Required number of ways = (8C5 x 10C6)

= (8C3 x 10C4)

=

8 x 7 x 6x

10 x 9 x 8 x 7

3 x 2 x 1 4 x 3 x 2 x 1

= 11760.

9. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls

be drawn from the box, if at least one black ball is to be included in the draw?



A. 32 B. 48

C. 64 D. 96

Answer: Option C

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

= 3 x6 x 5

+3 x 2

x 6 + 12 x 1 2 x 1

= (45 + 18 + 1)

= 64.

10. In how many different ways can the letters of the word 'DETAIL' be arranged in such a

way that the vowels occupy only the odd positions?

A. 32 B. 48

C. 36 D. 60

Answer: Option C

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

11. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men

and 3 women?

A. 63 B. 90

C. 126 D. 45



Answer: Option A

Explanation:

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =7 x 6

x 3 = 63.2 x 1

12. How many 4-letter words with or without meaning, can be formed out of the letters of the

word, 'LOGARITHMS', if repetition of letters is not allowed?

A. 40 B. 400

C. 5040 D. 2520

Answer: Option C

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= (10 x 9 x 8 x 7)

= 5040.

13. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so

that the vowels always come together?

A. 10080 B. 4989600

C. 120960 D. None of these

Answer: Option C

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the

rest are different.

Number of ways of arranging these letters =8!

= 10080.(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = 4! = 12.



2!

Required number of words = (10080 x 12) = 120960.

14. In how many different ways can the letters of the word 'OPTICAL' be arranged so that

the vowels always come together?

A. 120 B. 720

C. 4320 D. 2160

Answer: Option B

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.


15. If the best and the worst paper never appear together, then six examination papers can be

arranged in how many ways

A. 120 B. 240

C. 480 D. 2160

Answer: Option C

Explanation: If the best and the worst appear always together, the number of ways 5 ! * 2 .

Therefore required number of ways 6! - 5!*2 = 480.

16. In how many ways can 10 true-false questions be replied

A. 120 B. 720

C. 4320 D. 1024

Answer: Option D



Explanation: Required number of ways are 2¹⁰ = 1024 , because every question may be

answered in 2 ways.

17. 4 buses run between Bhopal and Gwalior. If a man goes from Gwalior to Bhopal by a bus

and comes back to Gwalior by another bus, then the total possible ways are

A. 12 B. 72

C. 43 D. 21

Answer: Option A

Explanation: Since the man can go in 4 ways and can back in 3 ways. Therefore total number

of ways is ways.

18. Find the total number of 9 digit numbers which have all the digits different.

A. 9 * 9 ! B. 10 !

C. 9! D. None of these

Answer: Option A

Explanation: 10P9 - 9P8 = 9 * 9!

19. How many combinations of students are possible if the group is to consist of exactly

3 freshmen?

A. 5000

B. 4550

C. 4000

D. 3550

Answer – (B)

Solution:

Here we need the number of possible combinations of 3 out of 5 freshmen,5C3,

and the number of possible combinations of 3 out of the 15 sophomores and

juniors, 15C3.



Note that we want 3 freshmen and 3 students from the other classes.

Therefore, we multiply the number of possible groups of 3 of the 5 freshmen

times the number of possible groups of 3 of the 15 students from the other

classes.

5C3×15C3= 4,550

20. How many combinations of students are possible if the group is to consist of exactly

3 freshmen and 3 sophomores?

A. 360

B. 460

C. 560

D. 660

Answer – (C)

Solution:

This second part of the problem is similar to the first except that the choice of

the second group of 3 comes only from the sophomores.

Again we want 3 freshmen and 3 sophomores so we multiply the number of

groups of freshmen times the number of groups of sophomores.

5C3×8C3= 560

21. How many combinations of students are possible if the group is to consist of an

equal number of freshmen, sophomores, and juniors?

A. 5880

B. 6880

C. 7880

D. 8880

Answer – (A)

Solution:

An equal number of students from each of the three classes mean 2 students

from each class, that is, 2 freshmen and 2 sophomores and 2 juniors.

5C2×8C2×7C2= 5,880

22. How many combinations of students are possible if the group is to consist of all

members of the same class?



A. 20

B. 25

C. 30

D. 35

Answer – (D)

Solution:

This part of the problem seems similar to the first three parts but it is very

different.

Here we want 6 freshmen or 6 sophomores or 6 juniors.

The group cannot be all freshmen since there are only 5 freshmen.

Therefore, the group can be a group of 6 sophomores or 6 juniors.

Number of groups of sophomores =8C6=28

Number of groups of juniors=7C6=7

Since we want the number of groups of 6 sophomores or 6 juniors, we want the

sum of each of these possibilities:

=28+7= 35

23. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all

the prizes?

A. 256

B. 24

C. 12

D. None of these

Answer – (B)

Solution:

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4.

Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)

B2 will receive prize from rest 3 available prizes(so 3 ways)

B3 will receive his prize from the rest 2 prizes available(so 2 ways)

Finally B4 will receive the the last remaining prize (so 1 way)

So total ways would be: 4*3*2*1=24 Ways

Hence, the 4 prizes can be distributed in 24 ways.

24. The letter of the word LABOUR are permuted in all possible ways and the words

thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

A. 275



B. 251

C. 240

D. 242

Answer – (D)

Solution:

The order of each letter in the dictionary is ABLORU.

Now, with A in the beginning, the remaining letters can be permuted in 5!

ways.

Similarly, with B in the beginning, the remaining letters can be permuted in 5!

ways.

With L in the beginning, the first word will be LABORU, the second will be

LABOUR.

Hence, the rank of the word LABOUR is 5!+5!+2= 242

25. There are fourteen juniors and twenty-three seniors in the Service Club. The club is

to send four representatives to the State Conference.

(i)- How many different ways are there to select a group of four students to attend

the conference?

(ii)- If the members of the club decide to send two juniors and two seniors, how

many different groupings are possible?

A. (i): 37C4 , (ii): 14C2×23C2

B. (i): 37P4 , (ii): 14P2×23P2

C. (i): 37P4 , (ii): 14C2×23C2

D. (i): 37C4 , (ii): 14P2×23P2

Answer – (A)

Solution:

(i) - Part (i) of the question can be solved by choosing 4 students from the total

number of students.

Order is not important.

37C4=66,045

(ii) – For part (ii) of the problem choose 2 juniors and 2 seniors. ⇒ 14C2×23C2=91×253

=23,023

26. A class photograph has to be taken. The front row consists of 6 girls who are sitting.

20 boys are standing behind. The two corner positions are reserved for the 2 tallest



boys. In how many ways can the students be arranged?

A. 6! × 1440

B. 18! × 1440

C. 18! ×2! × 1440

D. None of these

Answer – (B)

Solution:

Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18!

ways.

Girls can be arranged in 6! ways.

Total number of ways in which all the students can be arranged =2!×18!×6!

= 18! ×1440

27. If 5×nP3=4×n+1P3 find n?

A. 10

B. 12

C. 11

D. 14

Answer – (D)

Solution:

nP3=n(n–1)(n–2)

n+1P3=(n+1)n(n–1) [Replacing n by (n + 1)]⇒ 5n(n−1)(n−2)=4(n+1)(n)(n−1)

Divide both sides by n (n-1)

5(n−2)=4(n+1)

5n−10=4n+4

5n−4n=4+10

n= 14

28. Ten different letters of alphabet are given, words with 5 letters are formed from

these given letters. Then, the number of words which have at least one letter repeated

is:

A. 69760

B. 30240

C. 99748

D. 42386



Answer – (A)

Solution:

Number of words which have at least one letter replaced:

= Total number of words - total number of words in which no letter is repeated⇒105–16P5 ⇒100000−30240= 69760

29. 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in

these chairs so that the chairs numbered 1 to 8 should be occupied and no two men

occupy adjacent chairs. Find the number of ways the task can be done.

A. 360

B. 384

C. 432

D. 470

Answer – (B)

Solution:

Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be

occupied.

_X_ , 2,3,4,5,6,7,_X_,9,10,11,12.

The various combinations of chairs that ensure that no two men are sitting

together are listed.

(1,3,5,__). The fourth chair can be 5,6,10,11 or 12, hence 5 ways.

(1,4,8,__), The fourth chair can be 6,10,11 or 12 hence 4 ways.

(1,5,8,__), the fourth chair can be 10,11 or 12 hence 3 ways.

(1,6,8,__), the fourth chair can be 10,11 or 12 hence 3 ways.

(1,8,10,12) is also one of the combinations.

Hence, 16 such combinations exist.

In case of each these combinations we can make the four men inter arrange in

4! ways.

Hence, the required result =16×4!= 384

30. How many words can be formed by re-arranging the letters of the word ASCENT

such that A and T occupy the first and last position respectively?

A. 5!

B. 4!



C. 6!–2!

D. 6!

Answer – (B)

Solution:

As A and T should occupy the first and last position, the first and last position

can be filled in only one following way.

A _ _ _ _ T

The remaining 4 positions can be filled in 4! Ways by the remaining words

(S,C,E,N,T).

Hence by rearranging the letters of the word ASCENT we can form:

1x4!= 4! Words.

31. If 6Pr=360 and 6Cr=15 find r ?

A. 4

B. 3

C. 6

D. 5

Answer – (A)

Solution:

nPr=nCr×r!

6Pr=15×r!

360=15×r!

r!=

360

15

=24

=4×3×2×1⇒ r!=4!

Therefore r= 4

32. In how many ways can six different rings be worn on four fingers of one hand?

A. 10

B. 12

C. 15

D. 16

Answer – (C)



Solution:

Required number of ways =6C4=

6×5

2

= 15 ways.

33. There are 12 yes or no questions. How many ways can these be answered?

A. 4096

B. 2048

C. 1024

D. 144

Answer – (A)

Solution:

Each of the questions can be answered in 2 ways (yes or no)

Therefore, no. of ways of answering 12 questions =212= 4096 ways.

34. There are 7 non-collinear points. How many triangles can be drawn by joining these

points?

A. 35

B. 10

C. 8

D. 7

Answer – (A)

Solution:

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points.

The number of triangles formed =7C3

35. From 6 men and 4 ladies, a committee of 5 is to be formed. In how many ways can

this be done, if the committee is to include at least one lady?



A. 246

B. 340

C. 290

D. 315

Answer – (A)

Solution:

To committee can be formed in the following ways:

1 lady + 4 gents or 2 ladies + 3 gents or 3 ladies + 2 gents or 4 ladies + 1 gent or

5 ladies + 0 gents.

Total number of possible arrangements:

(4C1×6C4)+(4C2×6C3)+(4C3×6C2)+(4C4×6C1)⇒ 60+120+60+6= 246

36. A question paper consists of three sections 4,5 and 6 questions respectively.

Attempting one question from each section is compulsory but a candidate need not

attempt all the questions. In how many ways can a candidate attempt the questions?

A. 209

B. (4!-1)(5!-1)(6!-1)

C. 119

D. 29295

Answer – (D)

Solution:

At least 1 question from each section is compulsory, so from the 1st section the

candidate can attempt 1 or 2 or 3 or 4 questions.

In each section each question can be dealt with in 2 ways, i.e. either he attempts

it or leaves it.

So far 4 question there are 2(2)(2)(2) ways to attempt.

As he has to attempt at least 1 question, the total number of ways in which he

can attempt questions from 1st section is 24−1

Similarly for the 2nd section there are 25−1 ways in which he can attempt and

for the 3rd section there are 26−1 ways.

The ways in which the attempts one or more questions in any section is

independent of the number of ways in which he attempts one or more questions

from the other sections.

Thus, total number of ways in which he can attempt questions in that paper:

=(24−1)(25−1)(26−1)=15×31×63= 29295



37. In how many ways a President, VP and Waterboy can be selected from a group of 10

people.

A. 10C3

B. 10P3

C. 240

D. 360

Answer – (B)

Solution:

We are selecting three different posts here, so order matters.

Thus total ways of selecting a President, VP and Waterboy from a group of 10

people would be 10P3

38. In a hockey championship, there are 153 matches played. Every two team played

one match with each other. The number of teams participating in the championship

is:

A. 18

B. 19

C. 17

D. 16

Answer – (A)

Solution:

Let there were x teams participating in the games, then total number of matches:

=nC2=153

On solving we get n=−17 and n=18.

It cannot be negative so n= 18 is the answer.

39. A box contains 10 balls out of which 3 are red and rest are blue. In how many ways

can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls

are included in the sample and no sample has all the 6 balls of the same colour?

A. 105

B. 168

C. 189

D. 120

Answer – (B)



Solution:

Six balls can be selected in the following ways:

One red ball and 5 blue ball

Or

Two red balls and 4 blue balls

Total number of ways:

=(3C1×7C5)+(3C2×7C4)=63+105= 168

40. Out of eight crew members three particular members can sit only on the left side.

Another two particular members can sit only on the right side. Find the number of

ways in which the crew can be arranged so that four men can sit on each side.

A. 864

B. 863

C. 865

D. 1728

Answer – (D)

Solution:

Required number of ways =3C2×4!×4!= 1728

41. A man positioned at the origin of the coordinate system. the man can take steps of

unit measure in the direction North, East, West or South. Find the number of ways of

he can reach the point (5,6), covering the shortest possible distance.

A. 252

B. 432

C. 462

D. 504

Answer- (C)

Solution:

In order to reach (5,6) covering the shortest distance at the same time the man

has to make 5 horizontal and 6 vertical steps.

The number of ways in which these steps can be taken is given by:

42. There are 6 equally spaced points A,B,C,D,E and F marked on a circle with radius R.



How many convex pentagons of distinctly different areas can be drawn using these

points as vertices?

A. 6P5

B. 1

C. 5

D. None of these

Answer – (D)

Solution:

Since, all the points are equally spaced; hence the area of all the convex

pentagons will be same.

43. In an examination paper, there are two groups each containing 4 questions. A

candidate is required to attempt 5 questions but not more than 3 questions from any

group. In how many ways can 5 questions be selected?

A. 24

B. 48

C. 96

D. 64

Answer – (B)

Solution:

5 questions can be selected in the following ways:

2 question from first group and 3 question from second group

Or

3 question from first group and 2 questions from second group.

(4C2×4C3)+(4C3×4C2)=24+24= 48

44. After every get-together every person present shakes the hand of every other person.

If there were 105 handshakes in all, how many persons were present in the party?

A. 14

B. 13

C. 15

D. 16

Answer – (C)

Solution:

Let total number of persons present in the party be x,

Then



x= 15

45 How many diagonals can be drawn in a pentagon?

A. 5

B. 10

C. 8

D. 7

Answer – (A)

Solution:

A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs.

Total number of sides and diagonals:

This includes its 5 sides also.⇒ Diagonals =10–5=5

Hence the number of diagonals=10–5= 5

46. There are five women and six men in a group. From this group a committee of 4 is

to be chosen. How many different ways can a committee be formed that contain

three women and one man?

A. 55

B. 60

C. 25

D. 192

Answer – (B)

Solution:

Since no order to the committee is mentioned, a combination instead of a

permutation is used.

Let’s sort out what we have and what we want.

Have: 5 women, 6 men.

Want: 3 women AND 1 man.

The word AND means multiply.

Woman and Men



haveCwant × haveCwant

5C3×6C1= 60

47

.

Find the total number of distinct vehicle numbers that can be formed using

two letters followed by two numbers. Letters need to be distinct.

A. 60000

B. 65000

C. 70000

D. 75000

Answer – (B)

Solution:

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways.

Coming to numbers part, there are 10 ways (any number from 0 to 9 can be

chosen) to choose the first digit and similarly another 10 ways to choose the

second digit.

Hence there are totally 10×10=100 ways.

Combined with letters there are 26P2×100= 65000 ways to choose vehicle

numbers.

48. In a railway compartment, there are 2 rows of seats facing each other with

accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the

rear while 3 others are indifferent. In how many ways can the 10 passengers be

seated?

A. 172000

B. 12600

C. 45920

D. 43200

Answer – (D)

Solution:

The four person who wish to sit facing forward can be seated in: 5P4 ways and

3 who wish to sit facing towards the rear can be seated in: 5P3 ways and the

remaining 3 can be seated in the remaining 3 seats in 3P3 ways.

Total number of ways =5P4×5P3×3P3= 43200

49. There are three prizes to be distributed among five students. If no students gets more

than one prize, then this can be done in:



A. 10 ways

B. 30 ways

C. 60 ways

D. 80 ways

Answer – (A)

Solution:

3 prize among 5 students can be distributed in 5C3 ways = 10 ways.

50. A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can,

without taking the same pair of children together more than once. How many times

does the teacher go to the zoo?

A. 10

B. 12

C. 15

D. 20

Answer – (C)

Solution:

Two students can be selected from 6 in 6C2=15 ways.

Therefore, the teacher goes to the zoo 15 times.

51. A family consist of a grandfather, 5 sons and daughter and 8 grandchildren. They are

to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each

end and the grandfather refuses to have a grandchild on either side of him. The

number of ways in which the family can be made to sit is:

A. 21530

B. 8! × 480

C. 8! × 360

D. 8! × 240

Answer – (B)

Solution:

Total no. of seats:

= 1 grandfather+ 5 sons and daughters + 8 grandchildren = 14.

The grandchildren can occupy the 4 seats on either side of the table in 4! = 24

ways.

The grandfather can occupy a seat in (5-1)= 4 ways (4 gaps between 5 sons and

daughter).



And, the remaining seats can be occupied in 5!= 120 ways (5 seat for sons and

daughter).

Hence total number of required ways = 8! × 480

TWO MARK QUESTIONS

1. If the letters of the word CHASM are rearranged to form 5 letter words such that

none of the word repeat and the results arranged in ascending order as in a dictionary

what is the rank of the word CHASM?

A. 24

B. 31

C. 32

D. 30

Answer – (C)

Solution:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters

repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.

The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that

start with CA.

The next word starts with CH and then A, i.e., CHA _ _.

The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

2. In how many ways can 5 different toys be packed in 3 identical boxes such that no

box is empty, if any of the boxes may hold all of the toys?

A. 20

B. 30

C. 25

D. 600

E. 480



Answer – (C)

Solution:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the

following ways

a. 2,2,1

b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the

remaining 3 can be selected in 3C2 ways and the last toy can be selected

in1C1 way.

However, as the boxes are identical, the two different ways of selecting which

box holds the first two toys and which one holds the second set of two toys will

look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in5C3 ways. As the boxes are identical,

the remaining two toys can go into the two identical looking boxes in only one

way.

Therefore, total number of ways of getting the 3−1–1 option is 5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

=number of ways of achieving Case a + number of ways of achieving Case b

=15+10= 25 ways.

3. What is the value of 1×1!+2×2!+3!×3!+............n×n!;

where n! means n factorial or n(n−1)(n−2)...1

A. n(n−1)(n−1)!

B. (n+1)!

n(n−1)

C. (n+1)!−n!

D. (n+1)!−1!

Answer – (D)

Solution:

1×1!=(2−1)×1!=2×1!−1×1!=2!−1!



2×2!=(3−1)×2!=3×2!−2!=3!−2!

3×3!=(4−1)×3!=4×3!−3!=4!−3!

..

n×n!=(n+1−1)×n!=(n+1)(n!)−n!=(n+1)!−n!

Summing up all these terms, we get (n+1)! - 1!

4. When six fair coins are tossed simultaneously, in how many of the outcomes will at

most three of the coins turn up as heads?

A. 25

B. 41

C. 22

D. 42

E. 31

Answer – (D)

Solution:

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is 6C0=1 outcomeThe number of outcomes in which 1 coin turns head is 6C1=6 outcomesThe number of outcomes in which 2 coins turn heads is 6C2=15 outcomesThe number of outcomes in which 3 coins turn heads is 6C0=20 outcomes.

Therefore, total number of outcomes =1+6+15+20= 42 outcomes.

5. A college has 10 basketball players. A 5-member team and a captain will be selected

out of these 10 players. How many different selections can be made?

A. 1260

B. 210

C. 10C6×6!

D. 10C5×6

Answer – (A)

Solution:

A team of 6 members has to be selected from the 10 players. This can be done

in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 210×6= 1260

Alternatively, we can select the 5 member team out of the 10

in 10C5 ways=252 ways.REAVATHY.N, ASST PROF, MATHS DEPT


The captain can be selected from amongst the remaining 5 players in 5 ways.

Therefore, total ways the selection of 5 players and a captain can be made

=252×5= 1260

6. A 6x6 grid is cut from an 8x8 chessboard. In how many ways can we put two

identical coins, one on the black square and one on a white square on the grid, such

that they are not placed in the same row or in the same column?

A. 216

B. 324

C. 144

D. 108

Answer – (A)

Solution:

In a 6x6 grid of a chessboard, each row and each column contains 3 white and 3

black squares placed alternatively.

There are a total of 18 black and 18 white squares.

For every black square chosen to put one coin, we cannot choose any white

square present in its row or column.

There are 3 white squares in its row and 3 white square in its column for every

black square.

Hence for every black square chosen, we can choose (18−6)=12 white squares.

Total number of possibilities where a black square and a white square can be

chosen so that they do not fall in the same row or in the same column:

=18×12=216

So there are 216 ways of placing the coins that are identical.

7. How many four letter distinct initials can be formed using the alphabets of English

language such that the last of the four words is always a consonant?

A. 263×21

B. 26×25×24×21

C. 25×24×23×21

D. None of these

Answer – (A)

Solution:

The last of the four letter words should be a consonant. Therefore, there are 21

options.



The first three letters can be either consonants or vowels. So, each of them have

26 options. Note that the question asks you to find out the number of distinct

initials and not initials where the letters are distinct.

Hence answer =26×26×26×21=263×21

8. What is the total number of ways in which Dishu can distribute 9 distinct gifts

among his 8 distinct girlfriends such that each of them gets at least one gift?

A. 72 × 8!

B. 144 × 8!

C. 36 × 8!

D. 9!

Answer – (C)

Solution:

As every girl friend should get one gift.

The number of ways 8 distinct gifts can be selected is: 9C8=9 ways.

The number of ways each GF gets one gift each out of these 8 selected gifts 8!

Total number of ways 8 gifts can be distributed is 9×8!.

Now the last one gift can be given to any of the 8 GF hence the total number of

ways of distributing

9. How many number of times will the digit ‘7' be written when listing the integers

from 1 to 1000?

A. 271

B. 300

C. 252

D. 304

Answer – (B)

Solution:

7 does not occur in 1000. So we have to count the number of times it appears

between 1 and 999. Any number between 1 and 999 can be expressed in the

form of xyz where 0≤x,y,z≤9.

1. The numbers in which 7 occurs only once. e.g 7,17,78,217,743 etc.

This means that 7 is one of the digits and the remaining two digits will be any

of the other 9 digits (i.e, 0 to 9 with the exception of 7)

You have 1×9×9=81 such numbers.



However, 7 could appear as the first or the second or the third digit.

Therefore, there will be 3×81=243 numbers (1-digit, 2-digits and 3- digits) in

which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to

9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in

the first or second or the third place. So there are 3×9=27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written

54 times.

3. The number in which 7 appears thrice: 777–1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is:

243+54+3= 300

10. In how many ways can 15 people be seated around two round tables with seating

capacities of 7 and 8 people?

A. 15!

8!

B. 7!×8!

C. 15C8×6!×7!

D. 2×15C7×6!×7!

E. 15C8×8!

Answer – (C)

Solution:

Circular Permutation

'n' objects can be arranged around a circle in (n−1)!

If arranging these 'n' objects clockwise or counter clockwise means one and the

same, then the number arrangements will be half that number.

i.e., number of

You can choose the 7 people to sit in the first table in 15C7 ways.

After selecting 7 people for the table that can seat 7 people, they can be seated

in:(7−1)!=6!

The remaining 8 people can be made to sit around the second circular table in:

(8−1)!=7! Ways.



Hence, total number of ways: 15C8×6!×7!

11. In how many ways can the letters of the word EDUCATION be rearranged so that

the relative position of the vowels and consonants remain the same as in the word

EDUCATION?

A. 9!

4

B. 9!

4!×5!

C. 4!×5!

D. None of these

Answer – (C)

Solution:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5

positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement

should remain the same as in the word EDUCATION, the vowels can occupy

only the afore mentioned 4 places and the consonants can

occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in 1st,2nd,4th,6th and 9th position

in5! Ways.

Hence, the total number of ways = 4! × 5!

12. A committee is to be formed comprising 7 members such that there is a simple

majority of men and at least 1 woman. The shortlist consists of 9 men and 6 women.

In how many ways can this committee be formed?

A. 3724

B. 3630

C. 4914

D. 3824

Answer – (C)

Solution:

Three possibilities:- 1W+6M, 2W+5M, 3W+4M⇒ (6C1×9C6)+(6C2×9C5)+(6C3×9C4)=4914



13. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and

the other only 4. In how many ways can they travel?

A. 9

B. 26

C. 126

D. 392

Answer – (C)

Solution:

There are 8 students and the maximum capacity of the cars together is 9.We may divide the 8 students as followsCase I: 5 students in the first car and 3 in the second Or Case II: 4 students in the first car and 4 in the secondHence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.Therefore, the total number of ways in which 8 students can travel is:8C3+8C4=56+70= 126

14. A tea expert claims that he can easily find out whether milk or tea leaves were added

first to water just by tasting the cup of tea. In order to check this claims 10 cups of

tea are prepared, 5 in one way and 5 in other. Find the different possible ways of

presenting these 10 cups to the expert.

A. 252

B. 240

C. 300

D. 340

Answer – (A)

Solution:

Since there are 5 cups of each kind, prepared with milk or tea leaves added first,

are identical hence, total number of different people ways of presenting the cups

to the expert is

15. How many alphabets need to be there in a language if one were to make 1 million

distinct 3 digit initials using the alphabets of the language?

A. 100

B. 50

C. 26



D. 1000

Answer – (A)

Solution:

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n×n×n=n3 distinct 3 digit initials.

Note: Distinct initials are different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct

initials while they are not permitted when the digits of the initials need to be

different.

This n3 different initials = 1 million

i.e. n3=106 (1 million =106)⇒ n3=(102)3 ⇒ n=102=100

Hence, the language needs to have a minimum of 100 alphabets to achieve the

objective

16. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which

the first 1 occur before the first 2?

A. 420

B. 360

C. 320

D. 210

Answer – (D)

Solution:

Total number of arrangements

Since, there are two 1's and two 0's, the number of arrangements in which the

first 1 is before the first 2 is same as the number of arrangement in which the

first 2 is before the first 1 and they are each equal to half the total number of

arrangements = 210

17. How many different five-letter words can be formed using the letter from the world

APPLE?

A. 120

B. 60



C. 240

D. 24

Answer – (B)

Solution:

If the two P’s were distinct (they could have different subscripts and colours),

the number of possible permutations would have been 5!=120

For example let us consider one permutation: P1LEAP2

Now if we permute the P’s amongst them we still get the same word PLEAP.

The two P’s can be permuted amongst them in 2! ways.

We were counting P1LEAP2 and P2LEAP1 as different arrangements only

because we were artificially distinguishing between the two P’s Hence the

number of different five letter words that can be formed is:

18. I have an amount of Rs 10 lakh, which I went to invest in stocks of some companies.

I always invest only amounts that are multiples of Rs 1 lakh in the stock of any

company. If I can choose from among the stocks of five different companies, In how

many ways can I invest the entire amount that I have?

A. 252

B. 250

C. 1001

D. 1089

Answer – (C)

Solution:

The situation is similar to placing 10 identical balls among 5 distinguishable

boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5-1) four walls in the

single row, which can be done in 14C4 ways.(The balls placed between every

successive pair of walls belong to one group) 14C4= 1001 ways.

19. For a Lofoya Assets, a selection committee is to be chosen consisting of 5 ex-

technicians. Now there are 12 representatives from four zones. It has further been



decided that if Mr. Lofs is selected, Relk and Lemini will not be selected and vice-

versa. In how many ways it can be done?

A. 572

B. 672

C. 472

D. 372

Answer – (B)

Solution:

10C5:- when both are not included.

10C4:- when one of them is included.

Number of ways =10C5+10C4+10C4= 672

20. How many 5-digit positive integers exist the sum of whose digits are odd?

A. 36000

B. 38000

C. 45000

D. 90000

Answer – (C)

Solution:

There are 9×104=90000 5-digit positive integers.

Out of these 90000 positive integers, the sum of the digits of half of the

numbers will add up to an odd number and the remaining half will add up to an

even number.

Hence, there are 5-digit positive integers whose sum add up

to an odd number.

21. There are eight boxes of chocolates, each box containing distinct number of

chocolates from 1 to 8. In how many ways four of these boxes can be given to four

persons (one boxes to each) such that the first person gets more chocolates than each

of the three, the second person gets more chocolates than the third as well as the

fourth persons and the third person gets more chocolates than fourth person?

A. 35

B. 70

C. 105

D. 210



Answer – (B)

Solution:

All the boxes contain distinct number of chocolates.

For each combination of 4 out of 8 boxes, the box with the greatest number has

to be given to the first person, the box with the second highest to the second

person and so on.

The number of ways of giving 4 boxes to the 4 person is: 8C4= 70

22. In how many ways can seven friends be seated in a row having 35 seats, such that no

two friends occupy adjacent seats?

A. 29P7

B. 29C7

C. 28P7

D. 29C7

Answer – (A)

Solution:

First let us consider the 28 unoccupied seats.

They create 29 slots- one on the left of each seat and one on the right of the last

one.

We can place the 7 friends in any of these 29 slots i.e. 29P7 ways.

23. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes

can take more than 10 letters?

A. 510

B. 105

C. 10P5

D. 10C5

Answer – (A)

Solution:

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth

for all of the 10 letters.

i.e.5×5×5×….×5 (up-to 10 times)

=510

24. There are 2 brothers among a group of 20 persons. In how many ways can the group

be arranged around a circle so that there is exactly one person between the two



brothers?

A. 2 × 17!

B. 18! × 18

C. 19! × 18

D. 2 × 18!

Answer – (D)

Solution:

Circular Permutation

'n' objects can be arranged around a circle in (n−1)!

If arranging these 'n' objects clockwise or counter clockwise means one and the

same, then the number arrangements will be half that number.

i.e., number of

Let there be exactly one person between the two brothers as stated in the

question.

If we consider the two brothers and the person in between the brothers as a

block, then there will 17 others and this block of three people to be arranged

around a circle.

The number of ways of arranging 18 objects around a circle is in 17! ways.

Now the brothers can be arranged on either side of the person who is in between

the brothers in 2! ways.

The person who sits in between the two brothers could be any of the 18 in the

group and can be selected in 18 ways.

Therefore, the total number of ways 18×17!×2= 2 × 18!

25. a,b,c,d and e are five natural numbers. Find the number of ordered sets (a,b,c,d,e)

possible such that a+b+c+d+e=64.

A. 64C5

B. 63C4

C. 65C4

D. 63C5

Answer – (B)

Solution:

Let assume that there are 64 identical balls which are to be arranged in 5

different compartments (Since a,b,c,d,e are distinguishable) If the balls are

arranged in a row



i.e., o,o,o,o,o,o......(64 balls)

We have 63 gaps where we can place a wall in each gap, since we need 5

compartments we need to place only 4 walls.

We can do this in 63C4 ways.

26. There are five cards lying on the table in one row. Five numbers from among 1 to

100 have to be written on them, one number per card, such that the difference

between the numbers on any two adjacent cards is not divisible by 4. The remainder

when each of the 5 numbers is divided by 4 is written down on another card (the 6th

card) in order. How many sequences can be written down on the 6th card?

A. 210

B. 210×33

C. 4×34

D. 4233

Answer – (C)

Solution:

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.

The remainder of the number on the next card when divided by 4 can have 3

possible vales (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: 4×34

27. From a total of six men and four ladies a committee of three is to be formed. If Mrs.

X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is

willing to join the committee only if Mrs Z is included, how many such committee

are possible?

A. 138

B. 128

C. 112

D. 91

Answer – (D)

Solution:

We first count the number of committee in which

(i). Mr. Y is a member

(ii). the ones in which he is not

case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.



Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+2C1=7$ ways.

case (ii): If Mr. Y is not a member then we left with (6+4-1) people.

we can select 3 from 9 in 9C3=84 ways.

Thus, total number of ways is 7+84= 91 ways.

28. Goldenrod and No Hope are in a horse race with 6 contestants. How many different

arrangements of finishes are there if No Hope always finishes before Goldenrod and

if all of the horses finish the race?

A. 700

B. 360

C. 120

D. 24

E. 21

Answer – (B)

Solution:

Two horses A and B, in a race of 6 horses... A has to finish before B

If A finishes 1... B could be in any of other 5 positions in 5 ways and other

horses finish in 4! Ways, so total ways =5×4!

If A finishes 2... B could be in any of the last 4 positions in 4 ways. But the

other positions could be filled in 4! ways, so the total ways =4×4!

If A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other

positions could be filled in 4! ways, so total ways =3×4!

If A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other

positions could be filled in 4! ways, so total ways =2×4!

If A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4!

ways.

A cannot finish 6th, since he has to be ahead of B.

Therefore total number of ways:

5×4!+4×4!+3×4!+2×4!+4!=120+96+72+48+24= 360

29. Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy

Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per

plant respectively. If he has to buy at least one of each plant and cannot buy any

other type of plants, then in how many distinct ways can Jay make his purchase?

A. 2



B. 3

C. 4

D. 5

Answer – (B)

Solution:

Let the number of Rose plants be ‘a’.

Let number of marigold plants be ‘b’.

Let the number of Sunflower plants be ‘c’.⇒ 20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,

19a+4b=900

b being the number of plants, is a positive integer, and is less than 99, as each of

the other two types have at least one plant in the combination i.e .:

0<b<99 ----------(2)

Substituting (1) in (2),

a is the integer between 47 and 27 ----------(3)

From (1), it is clear, a should be multiple of 4.

hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100

For all other values of a, we get the desired solution:

a=36,b=54,c=10

a=40,b=35,c=25

a=44,b=16,c=40

Three solutions are possible.

30. How many words of 4 consonants and 3 vowels can be made from 12 consonants

and 4 vowels, if all the letters are different?

A. 16C7×7!

B. 12C4×4C3×7!



C. 12C3×4C4

D. 12C4×4C3

Answer – (B)

Solution:

4 consonants out of 12 can be selected in 12C4 ways.

3 vowels can be selected in 4C3 ways.

Therefore, total number of groups each containing 4 consonants and 3

vowels =12C4×4C3

Each group contains 7 letters, which can be arranging in 7! ways.

Therefore required number of words =12C4×4C3×7!

31. A local delivery company has three packages to deliver to three different homes. if

the packages are delivered at random to the three houses, how many ways are there

for at least one house to get the wrong package?

A. 3

B. 5

C. 3!

D. 5!

Answer – (B)

Solution:

The possible outcomes that satisfy the condition of "at least one house gets the

wrong package" are:

One house gets the wrong package or two houses get the wrong package or

three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach

this problem from a different angle.

The only case which is left out of the condition is the case where no wrong

packages are delivered.

If we determine the total number of ways the three packages can be delivered

and then subtract the one case from it, the remainder will be the three cases

above.

There is only one way for no wrong packages delivered to occur.

This is the same as everyone gets the right package.

The first person must get the correct package and the second person must get

the correct package and the third person must get the correct package.⇒ 1×1×1=1



Determine the total number of ways the three packages can be delivered.⇒ 3×2×1=6

The number of ways at least one house gets the wrong package is:⇒ 6−1=5

Therefore there are 5 ways for at least one house to get the wrong package.

32. How many natural numbers less than a lakh can be formed with the digits 0,6 and 9?

A. 242

B. 243

C. 728

D. 729

Answer – (A)

Solution:

The digits to be used are 0,6 and 9.

The required numbers are from 1 to 99999.

The numbers are five digit numbers.

Therefore, every place can be filled by 0,6 and 9 in 3 ways.

Total number of ways =3×3×3×3×3=35

But 00000 is also a number formed and has to be excluded.

Total number of numbers =35−1=243−1= 242

33. There are 20 couples in a party. Every person greets every person except his or her

spouse. People of the same sex shake hands and those of opposite sex greet each

other with a namaskar (It means bringing one's own palms together and raising them

to the chest level). What is the total number of handshakes and namaskar's in the

party?

A. 760

B. 1140

C. 780

D. 720

Answer – (B)

Solution:

There are 20 men and 20 women. When a man meets a woman, there are two

namaskars, whereas when a man meets a man (or a woman) there is only 1



handshake. Number of handshakes =2×20C2 (men and women ) =2*380=760

For number of Namskars Every man does 19 namaskars (to the 20 women

excluding his wife) and they respond in the same way.

Number of Namaskars =(20)(19)=380

Total =760+380= 1140

34. In how many ways can a leap year have 53 Sundays?

A. 365C7

B. 7

C. 4

D. 2

Answer – (D)

Solution:

In a leap year there are 366 days i.e. 52 weeks + 2 extra days.

So to have 53 Sundays one of these two days must be a Sunday.

This can occur in only 2 ways.

i.e. (Saturday and Sunday) or (Sunday and Monday).

Thus number of ways =2

35. In a certain laboratory, chemicals are identified by a colour-coding system. There are

20 different chemicals. Each one is coded with either a single colour or a unique

two-colour pair. If the order of colours in the pairs does not matter, what is the

minimum number of different colours needed to code all 20 chemicals with either a

single colour or a unique pair of colours?

A. 7

B. 6

C. 5

D. 8

Answer – (B)

Solution:

Each one coded with either a single colour or unique two-colour pair.

Therefore, total number of ways =n+nC2

Minimum number of different colour needed to code all 20 chemicals will be 6.

=6+6C2=21

36. Six boxes are numbered 1,2,34,5 and 6. Each box must contain either a white ball or



a black ball. At least one box must contain a black ball and boxes containing black

balls must be consecutively numbered. find the total number of ways of placing the

balls.

A. 15

B. 20

C. 21

D. 36

Answer – (C)

Solution:

If there is 1 black ball, it can be placed in 6 ways.

If there are 2 black balls, they can be placed in 5 ways (in 1,2 ; 2,3 ; 3,4 ; 4,5

and 5,6) and so on.

If there are 6 black balls, they can be placed in 1 way.

The total number of ways of placing the balls is 1+2+3+4+5+6= 21

37. In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular

red toys are never together whereas 2 particular green toys are always together?

A. 11! × 2!

B. 9! × 90

C. 4 × 10!

D. 18 × 10!

Answer – (D)

Solution:

Considering two green toys that are to be together as one unit.

We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2

who are not to be together) is:

=9!×2!×10C2×2!= 18× 10!

38. There are five comics numbered from 1 to 5. In how many ways can they be

arranged, so that part-1 and part-3 are never together?

A. 48

B. 72

C. 120

D. 210

Answer – (B)

Solution:



The total number of ways in which 5 part can be arranged =5!=120.

The total number of ways in which part-1 and part-3 are always together:

=4!×2!=48

Therefore, the total number of arrangements, in which they are not together is:

=;120−48= 72

39. The number of ways which a mixed double tennis game can be arranged amongst 9

married couples if no husband and wife play in the same is:

A. 1514

B. 1512

C. 3024

D. 3028

Answer – (B)

Solution:

As per the question there are 9 married couples and no husband and wife should

play in the same game:

We know that in a mixed double match there are two males and two females.

Step I: Two male members can be selected in 9C2=36 ways

Step II: Having selected two male members, 2 female members can be selected

in 7C2=21 ways.

Step III: Two male and two female members can arranged in a particular game

in 2 ways.

Total number of arrangements=36×21×2= 1512 ways.

40. Ten coins are tossed simultaneously. In how many of the outcomes will the third

coin turn up a head?

A. 210

B. 29

C. 3×28

D. None of these

Answer – (B)

Solution:

When a coin is tossed once, there are two outcomes. It can turn up a head or a

tail.

When 10 coins are tossed simultaneously, the total number of outcomes=210

Out of these, if the third coin has to turn up a head, then the number of



possibilities for the third coin is only 1 as the outcome is fixed as head.

Therefore, the remaining 9 coins can turn up either a head or a tail=29

41. There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different

albums can be formed using the above repertoire if the albums should contain at

least 1 Rock song and 1 Carnatic song?

A. 15624

B. 16384

C. 6144

D. 240

Answer – (A)

Solution:

There are 2n ways of choosing ‘n’ objects.

For e.g. if n = 3, then the three objects can be chosen in the following23 ways:

3C0 ways of choosing none of the three,

3C1 ways of choosing one out of the three,

3C2 ways of choosing two out of the three and

3C3 ways of choosing all three.

In the given problem, there are 5 Rock songs. We can choose them in25 ways.

However, as the problem states that the case where you do not choose a Rock

song does not exist (at least one rock song has to be selected), it can be done in:

255−1=32−1=31 ways.

Similarly, the 6 Carnatic songs, choosing at least one, can be selected in:

26−1=64−1=63 ways.

And the 3 Indi pop can be selected in:

23=8 ways.

Here the option of not selecting even one Indi Pop is allowed.

Therefore, the total number of combinations=31×63×8= 15624

42. A book-shelf can accommodate 6 books from left to right. If 10 identical books on

each of the languages A,B,C and D are available, In how many ways can the book

shelf be filled such that book on the same languages are not put adjacently.

A. 40P6

6!

B. 10×95



C. 6P4

2!

D. 4×35

Answer – (D)

Solution:

First place can be filled in 4 ways.

The subsequent places can be filled in 3 ways each.

Hence, the number of ways =4×3×3×3×3×3=4×35

43. In how many ways can the letters of the word ABACUS be rearranged such that the

vowels always appear together?

A. 6!

B. 3!×3!

C. 4!

D.

E. 5!

Answer – (D)

Solution:

ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together as stated in the question, then there will

3 consonants and a set of 3 vowels grouped together.

One group of 3 vowels and 3 consonants are essentially 4 elements to be

rearranged.

The number of possible rearrangements is 4!

The group of 3 vowels contains two 'a's and one 'u'.The 3 vowels can rearrange

amongst themselves in ways as the vowel "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together

are:

44. If the letters of the word SACHIN are arranged in all possible ways and these words

are written out as in dictionary, then the word ‘SACHIN’ appears at serial number:

A. 601



B. 600

C. 603

D. 602

Answer – (A)

Solution:

If the word started with the letter A then the remaining 5 positions can be filled

in 5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways

Similarly if it started with H, I, N the remaining 5 positions can be filled in 5!

Ways

If it started with S then the remaining position can be filled with

A,C,H,I,N in alphabetical order as on dictionary

The required word SACHIN can be obtained after the 5×5!=600 Ways

i.e. SACHIN is the 601th letter

45. In how many ways can 3 men and their wives be made stand in a line such that none

of the 3 men stand in a position that is ahead of his wife?

A. 6!

B. 5!

C.

D. 7!

Answer – (C)

Solution:

6 people can be made to stand in a line in 6! Ways.

However, the problem introduces a constraint that no man stands in a position

that is ahead of his wife.

For any 2 given positions out of the 6 occupied by a man and his wife, the pair

cannot re-arrange amongst themselves in 2! Ways as the wife has to be in a

position ahead of the man. Only one of the 2! Arrangements are allowed.

As there are 3 couples in the group, the total number of ways gets reduced by a

factor of (2!×2!×2!).Hence, the total number of ways

46. In a cricket match if a batsman score 0,1,2,34,or 6 runs of a ball, then find the

number or different sequences in which he can score exactly 30 runs of an over.



Assume that an over consists of only 6 balls and there were no extra and no run outs.

A. 86

B. 71

C. 56

D. 65

Answer – (B)

Solution:

47. In how many ways can the letters of the word MANAGEMENT be rearranged so

that the two As do not appear together?

A. 10! - 2!

B. 9! - 2!

C. 10! - 9!

D. None of these

Answer – (D)

Solution:

The word MANAGEMENT is a 10 letter word.

Normally, any 10 letter word can be rearranged in 10! ways.

However, as there are certain letters of the word repeating, we need to account

for those. In this case, the letters A, M, E and N repeat twice each.

Therefore, the number of ways in which the letters of the word

MANAGEMENT can be rearranged reduces to:

The problem requires us to find out the number of outcomes in which the two

As do not appear together.

The number of outcomes in which the two As appear together can be found out



by considering the two As as one single letter. Therefore, there will now be only

9 letters of which three of them E, N and M repeat twice. So these 9 letters with

3 of them repeating twice can be rearranged in: ways.

Therefore, the required answer in which the two As do not appear next to each

other:=Total number of outcomes - the number of outcomes in which the 2 A’s

appear together

48. How many numbers are there between 100 and 1000 such that at least one of their

digits is 6?

A. 200

B. 225

C. 252

D. 120

Answer – (C)

Solution:

numbers between 100 and 1000:

=900

numbers between 100 and 1000 which do not have digit 6 in any place:

=8×9×9=648

Unit digit could take any value of the 9 values (0 to 9, except 6)

Tens Digit could take any value of the 9 values (0 to 9, except 6)

Hundreds digit could take any value of the 8 values (1 to 9, except 6)

numbers between 100 and 1000 which have at least one digit as 6:

=900−648= 252

49. A student is required to answer 6 out of 10 questions divided into two groups each

containing 5 questions. He is not permitted to attempt more than 4 from each group.

In how many ways can he make the choice?

A. 210

B. 150

C. 100

D. 200

Answer – (D)



Solution:

Number of ways of choosing 6 from 10 =10C6=210

Number of ways of attempting more than 4 from a group:

=2×5C5×5C1=10⇒ Required number of ways =210−10= 200

50. While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants,

3 half-pants,6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a

pair of shoes, a choice of "lower wear" (either a pant or a half-pant), a choice of

"upper wear" (it could be a shirt or a sweater or both) and finally he may or may not

choose to wear a jacket. How many different outfits are possible?

A. 567

B. 1821

C. 743

D. 1701

Answer – (D)

Solution:

Number of ways a pair of shoes can be selected =3C1=3 ways.

Number of ways "lower wear" can be selected =(3+4)=7 ways

Number of ways "upper wear" can be selected =3+6+3×6=27 ways

Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st

jacket, 2nd jacket)⇒ Total number of different outfits =3×7×27×3= 1701 ways.


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