# Permutation & Combination Problems with Solutions for bank exams-:

Today, I am going to share with you to solve “**permutation & combination questions”**. This chapter talk about selection and arrangement of things which could be any numbers, persons,letters,alphabets,colors etc.The basic difference between permutation and combination is of order Permutation is basically called as a arrangement where order does matters.Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time.”**permutation & combination”** section in **competitive exams** like IBPS PO,IBPS Clerk,SBI,RBI,RRB PO and Clerk examinations. This is an one of the

“Important Aptitude Topic“. This is an Aptitude PDF download file. Anyone can download it for free of cost. This Aptitude questions is for improve your“Aptitude skills”.

Right now there is fierce competition among candidates, it’s hard to crack any exam without scoring really good in **Quantitative Aptitude**.

**Download PDF ratio and proportion problems(link given below post)**

1. At an election, three wards of a town are to be canvassed by 4, 5 and 8 men respectively. If there are

20 men, in how many ways they can be allotted to the different wards?

(a) 4!*5!*8!/20! (b) 1240 (c)20890 (d)20!/17! (e)not

2. In how many different ways can 4 boys and 3 girls be arranged in a row such that all boys stand

together and all girls stand together?

(a) 288 (b) 144 (c) 56 (d) 64 (e) NOT

3. Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2

men. In how many different ways can it be done?

(a) 24 (b)30 (c) 45 (d) 72 (e) 10

4. In how many different ways can the letters of the word LONELY be arranged, so that the vowels are

at the two ends?

(a) 120 (b) 720 (c) 360 (d) 48 (e) 60

5. While packing for a business trip Rajesh 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‘, a

choice of ‗upper wear‘ and finally he may or may not choose to wear a jacket. How many different outfits

are possible?

(a) 240 (b) 3! (c) 5*3! (d) 2!*3! (e) 378

6. The number of ways in which the letters of the word CAPSULE can be rearranged so that the even

places are always occupied by consonants is:

(a) 576 (b) 6! (c) 4!*2! (d) 3!*6 (e) NOT

7. The number of ways in which 20 different flowers of two colors can be set alternatively on a

necklace, there being 10 flowers of each color is:

(a) 20! (b)10! * 10! (c) 20!*10 (d) 10!*2 (e) NOT

8. Calculate the number of ways a mixed double tennis player can be arranged, if there are a total of 9

couples and no couple plays in the same game?

(a) 1244 (b) 1512 (c) 1516 (d) 1412 (e) 2342

9. In how many different ways can the letter s of the word ‗PRESENTEE‘ be arranged?

(a) 15120 (b) 9! (c) 9! /2 (d) 9! /3 (e) NOT

10. 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) 26² (b) 26³ (c) 26³*21 (d) 26² *21 (e) 26*21

11. Find the numbers of ways in which the first, second, fourth and seventh letter of word EXHIBITION

can be arranged?

(a) 4! (b) 4! – 1! (c)5!-1 (d)5! +2 (e) NOT

12. Out of 5 women and 4 men, a committee of three members is to be formed in such a way that atleast

one member is a woman. In how many different ways can this be done?

(a) 150 ways (b) 200 ways (c) 196 ways (d) 120 ways (e) 156 ways

13. In how many ways can the letters of the word ‗RIDDLED‘ can be arranged?

(a) 70 ways (b) 35 ways (c) 81 ways (d) 720 ways (e)5040 ways

14. 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) 72 (b)25 (c) 36 (d) 64 (e) NOT

15. How many words of 4 consonant and 3 vowels can be made from 12 consonants and 4 vowels, if all

the letters are different?

(a) 12C4*4C3*7! (b)12C3*7! (c) 14C2 *2! (d) 13P2*7! (e) NOT

16. In how many ways can seven friends be seated in a row having 40 seats, such that no to friends

occupy adjacent seats?

(a) 40P7 (b) 40C2 (c) 35C2 (d) 36P7 (e) 36P7*2!

17. Find the total number of distinct vehicle numbers that can be formed using two letters followed by

two numbers. Letters need to be different.

(a) 4500 ways (b) 54000 ways (c) 65000 ways (d) 12000 ways (e) 26!

18. How many words can be formed by rearranging the letters of the word ASCENT such that A and T

occupy the first and last position respectively?

(a) 4! (b) 6! (c) 5!*2 (d) 4!*2 (e) NOT

19. Each of the letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror.

They are called symmetric letters. The other letters in the alphabet are asymmetric letters. How many three

letter computer password can be made without repletion with atleast two symmetric letters?

(a) 12870 (b) 12! (c) 12! /2! (d) Data inadequate (e) NOT

20. How many integers, greater 999 but not greater than 4000, can be formed with the digit 0, 1, 2, 3 and

4, if repletion of digits is allowed?

(a) 346 (b) 376 (c) 456 (d) data inadequate (e) NOT

21. In how many rearrangement of the word AMAZED, is the letter ‗E‘ positioned in between the 2

‗A‘s?

(a) 60 (b) 120*2! (c) 720 (d) 120 (e) 150

22. In a group of students, there are 5 freshmen, 8 sophomores and 7 juniors in a football club. A group

of 6 students will be chosen to compete in a completion. How many combinations of participating students

are possible if the group has to consist of all exactly 3 freshmen?

(a) 4500 (b) 5650 (c) 7240 (d) 4550 (e) 4510

23. Find the number of ways in which 8064 can be resolved as the product of two factors?

(a) 48 (b) 12 (c) 30 (d) 28 (e) 24

24. In how many different ways the letters of the word ‘DETAIL‘ be arranged in such a way that the

vowels occupy only the odd positions?

(a) 5! (b) 36 ways (c) 4! (d) 5!*2! (e) 4!*2!

25. Which of the following is the value of r, if 6Pr = 360 and 6Cr = 15.

(a) 7 (b) 4 (c) 2 (d) 5 (e) NOT

26. A class photograph has to be taken. The frontrow 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) 18!*10 (b) 18!*10! (c) 18!*1440 (d) 10!*18 (e) NOT

27. If the letters of the word LABOUR are permuted in all possible ways and the words formed by the

activity are arranged in a dictionary, then calculate the rank of the word LABORU.

(a) 105 (b) 241 (c) 240 (d) 120 (e) 242

28. In IPL, 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) 35 (c) 20 (d) 19 (e) 16

29. Calculate the number of diagonals which can be drawn in a hexagon?

(a) 15 (b) 9 (c) 7 (d) 12 (e) 13

30. If there are two brothers among a group of 20 people. Calculate the number of ways in which the

group can be arranged in a line so that they are always together?

(a) 18! (b)18!*1! (c) 3*18! (d) 18! / 3! (e) 19! * 2

31. In how many different ways the word‘ REPEAT‘ be arranged such that first and last position is held

by R and T respectively?

(a) 24 ways (b) 120 ways (c) 60 ways (d) 40 ways (e) NOT

32. Calculate the number of ways in which 6 yellow balls and 6 red balls be arranged, such that each

color group is always together?

(a) 12! (b) 10!*2 (c) 2!*6!*6! (d) 15! (e) 12!*18!

33. Calculate the number of 5 digit positive numbers, sum of whose digits is odd.

(a) 45000 (b) 23000 (c) 41000 (d) 13000 (e) 35000

34. 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) 1000 (b) 10 (c) 1500 (d) 100 (e) NOT

35. When four dice are rolled simultaneously, in how many outcomes will atleast one of the dice shows

3?

(a) 120 (b) 24 (c) 671 (d) 501 (e) 162

36. How many words can be formed by using all letters of the word ‗INHIBITION‘ so that vowels

always come together?

(a) 10! (b) 3600 (c) 7200 (d)1800 (e) NOT

37. 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) 120 (b) 150 (c) 140 (d) 105 (e) NOT

38. How many 3 digit can be formed from the digits 2, 4, 6, 8, 5, 7 and 1 which are divisible by 5 and

none of the digits is repeated?

(a) 120 (b) 200 (c) 250 (d) 40 (e)30

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

(a) 210 (b) 25200 (c) 210*2! (d) 70 (e) NOT

40. In how many ways a committee, consisting of 5 women and 6 men can be formed from 8 men and 15

women?

(a) 8C2*2! (b)8C6 * 15C5 (c) 8C5*15C6 (d) cannot be determined (e) NOT

Direction: each of the questions below consists of a question and two statements numbered I and II given

below it. You have to decide whether the data provided in the statements are sufficient to answer the

question. Read both statements and give answer as

(a). if the data in statement I is alone sufficient to answer the question

(b). if the data in statement II is alone sufficient to answer the question

(c). if the data in both the statements together is necessary to answer the question

(d). if the data in either statements is sufficient to answer the question.

(e). if the data in both the statements is not sufficient to answer the question

41. In how many ways letter of the word can be arranged?

I. The word is RECONCILATION.

II. The total alphabet of the digit is 12.

42. The ratio of arrangement of two words is?

I. First word is PRACTICAL.

II. Second word can be arranged in 60 ways.

43. Find how many alphabets are there in the word that is arranged in 180 ways?

I. The word is LEADER

II. The word has 3 same alphabets.

44. What is the value of r?

I. 10Cr = 1.

II. nPr =2.

45. In how many ways a committee of 6 members can be selected?

I. If 3 male members should be included.

II. If only 4 female members is to be included.

46. In how many ways a batch of 10 candidates can be selected for test?

I. There are total 56 candidates who filled the form.

II. Female candidate selected is 16.

47. In how many ways English and Hindi books are arranged?

I. Two books on Hindi may not be together.

II. There are 21 English and 19 Hindi books.

48. What is the value of n?

I. nPr = 1 and nC1 = 4.

II. nC1 = 1.

49. In how many different ways total students can be arranged?

I. There are total 4 boys and 3 girls in the group.

II. Arrangement is made such that all boys stand together and all girls stand together.

50. What is the ratio of ways of arrangement of two different words?

I. The ways of arranging the two words are 45 and 95.

II. First word is TAROT.

SOLUTION AND EXPLANATION OF PERMUTATION AND COMBINATION

1. (a)

The no. of ways so that 20 men can be allotted = 20!/4!*5!*8!

As after each allotment men get reduced.

2. (a)

BBBB GGG

No. of ways in which all girls stand together and all boys stand together = 2!*4!*3! = 288.

3. (b)

No. of ways of selecting 1 woman and 2 men = 3C1 * 5C2 = 30

4. (d)

There are 6 spaces in which both ends is to be filled by vowels, to arrange vowels there are2! Ways

The left 4 places can be filled by remaining alphabets in 4! Ways

So total no. of ways= 2! * 4! = 48

5. (e)

The no. of ways in which different outfit are possible = 3C1(shoes)*7C1(pants and half-pants)*9C1(shirts

and sweater)*2C1(jackets) = 3*7*9*2 = 378

6. (a)

The no. of ways in which consonant occupies even place = 4!

The no. of ways in which odd places is filled is 4! , so total ways= 4! * 4! = 576

7. (b)

Total ways in arranging flowers of one color = 10!

Total ways in which flowers of another color is arranged = 10!

Therefore total ways to arrange whole flowers = 10! * 10!

8. (b)

No. of double tennis player =9C2 = 36

No. of ways of forming a mixed double tennis player so that no couple plays in the same game =7C2=21

Total ways = 2(it is said couple) * 21 * 36 =1512

9. (a)

Total no. of ways = 9! /4! (4 E is present) = 15120

10. (c)

No. of ways of selecting consonant in the last letter = 21

Other letters can be filled by either consonant or vowel so there are 26 ways

Therefore Total ways =26*26*26*21 = 26³*21

11. (a)

The letters are E, X, I and T

So total no. of ways = 4!

12. (b)

Case.1- woman-1 and men=2

No. of ways = 5(woman) * 4(men) * 3(men) = 60

Case.2- 2 woman and 1 man

No. of ways= 5(woman) *4(woman) * 4(man)= 80

Case.3- all are woman

No. of ways = 5*4*3 = 60

Total ways = 60+80+60 = 200

13. (b)

No. of ways to arrange the word RIDDLED = 7! / 3! = 35

14. (a)

Total no. of ways in which 5 parts of comic can be arranged = 5! =120

Total no. of ways in which part-1 and part-3 are always together= 4!*2! = 48

Therefore total no. of ways = 120-48 = 72

15. (a)

The no. of ways to have 4 consonants and 3 vowels = 12C4 * 4C3

But to have letters all different no. of ways = 12C4 * 4C3 * 7!

16. (d)

Total no. of seat left after occupying by seven friends = 40-7 = 35

Therefore there are total 36 alternate seats.

Hence seven friends can be seated in 36C7 ways so that no two friend occupy adjacent seat

17. (c)

No. of ways to select 2 distinct alphabets from 26 = 26P2

Also there are total 10 digits so to select one number we have 10 ways and two select another number we

have 10 ways

So there are total 10*10 = 100 ways to select numbers

Then combinations of letters and numbers = 26P2 *100 = 65000 ways

18. (a)

First and last letter is to be filled by A and T

Then only four letters are left

So there are 4! Ways to arrange the word.

19. (e)

Case.1- two symmetric and 1 asymmetric

No. of ways = 11C2*15C1= 825

Case.2- all symmetric

No. of ways = 11C3 = 165

Total ways to have 3 letter password = 3!*(825+165) = 5940

20. (b)

The first digit is 1000, a 4 digit number

And the last digit = 4000, the only 4 digit numbers to start with 4

There fore there are four digits in each integer, and first digit can be 1, 2 and 3

Second, third and fourth cab be 0, 1, 2, 3 and 4 i.e. 5 ways

So total ways = 3*5*5*5 = 375 +1 for 4000

21. (d)

The combination of both A and E is AAE, AEA & EAA

In each case only AEA have E in between which is one-third of all combinations

Total no. of ways of arranging the word is 6! / 2! = 360

Total ways to have E in between = 360/3 = 120

22. (d)

No. of ways of have exactly 3 freshmen = 5C3 * 15C3 = 4550

23. (e)

Total pairs is (1*064), (2*4032), (3*2688), (4*2016), (6*1344), (7*1152), (8*1008), (9*896),12*672), (14*

576), (16 *504), (18*448), (21*884), (24*336), (28*288), (32*252), (36*224), (42*192), (48*168),

(56*144), (63*128), (68*126), (72*112) and (84*96). = 24 ways

24. (b)

Vowels to occupy odd position, no. of ways= 3!

Other positions can fill in 3! Ways

Total no. of ways = 3!*3! = 36

25. (b)

6Pr= 360, 6! / (6-r)! = 360

6Cr = 15, 6!/(6-r)!*r! = 15

So r! = 360/15 = 24(4*3*2*1)

So r= 4!

26. (c)

Two tallest boys can be arranged in 2! Ways

And other in 18! Ways

And arrangement of girls = 6!

Total ways of arrangement = 18!*2!*6! = 18! * 1440

27. (b)

The word is LABOUR

To arrange the word alphabetically, we have first word as ABLORU

So to have A as first letter there are 5! = 120 ways

Then second is B as first letters (BALORU) no . Of ways = 5! =120

Next word will be LABORU so its position = 120+120+1 = 241

28. (a)

According to the question nC2 = 153

Or n! / (n-2)! * 2! = 153=> n*(n-1)/2 = 153

On solving we get n= 18

29. (b)

Total no. of diagonals in a hexagon= 6C2=15

But there are sis sides also, so total diagonals = 15-6 = 9

30. (e)

There are total 20 people

Sine two brother always stand together so total ways = 19!*2!

31. (a)

First and last letter is to be filled by R and T

Then only four letters are left

So there are 4! =24 no. of ways to arrange the word.

32. (c)

Considering yellow balls a group and red balls a group

So no. of ways = 2!* 6! * 6!

33. (a)

There are total 9*10^4 = 90000 5 digit numbers

In 90000 numbers half will be even and half will be odd

So there are 45000 odd numbers

34. (d)

To make 3 letters word using 1 million alphabets

We have x^3 = 1million

Then x^3 = 10^6

So x = 100

35. (c)

Total no. of outcomes when 4 dice is thrown is 6^4 = 1296

Total no. of ways when none of dice shows 3 = 5^4 = 625

Total ways in which atleast one of them is 3 in four dice = 1296-625= 671

36. (d)

Total no. of ways so that vowels always come together=

37. (a)

Case.1- men-3, women = 2

No. of ways= 7C3 * 6C2= 35*15=525

Case.2- men = 4 and women = 1

No. of ways= 7C4 * 6C1 = 210

Case.2- men= 5 and women = 0

No. of ways = 7C5 = 21

Total ways = 525+210+21 = 756

38. (e)

For three digit numbers to be divisible by 5 last digits should be 5

Remaining two spaces can be filled as6 and 5 ways

So total ways to have 3 digit = 6*5*1(5) = 30

39. (b)

No. of ways of selecting (3 consonants and 2 vowels) = 7C3 * 4C2 = 210

No. of groups each having 3 consonants and 2 vowels = 210

Each group contains 5 letter, so no. of ways to arrange 5 letter among themselves =5!

There no. of words = 210 * 5! = 25200

40. (b)

No. of ways to form a committee consisting of 5 women and 6 men = 15C5 * 8C6

41. (a)

From statement I, total no. of ways = 13! / (2!*2!*2!*2!)

From statement II, total ways =12!, but alphabet is not given so it may or may not contain identical letters

but we do not have confirm result by statement II.

42. (c)

From statement I, first word is PRACTICAL no. of ways to arrange = 9! / (2!*2!) = 90720

From statement II, second word arrangement = 60 ways

Ratio = 90720/60 = 1512:1.

43. (e)

From statement I, word is LEADER, arrangement =6! /2! = 360

From statement II, identical alphabet= 3

Both statementsare not sufficient to answer the question.

44. (e)

From statement I, 10Cr =1, either r= 1 or r= 9

From statement II, nPr = 2 two unknown value is given.

45. (e)

From statement I, total male member present = 3, both how many member should be there in the committee

is not given

From statement II, total female members to be included = 4, not any other information is given.

46. (e)

From statement I, total candidate who filled the form = 56 dat is not given regarding the selection of

candidates.

From statement II, female candidate= 16 no other information is given which hirs needed to answer the

question

47. (b)

From statement I, no. of English books is not given

From statement II, there is (21+19)! Ways to arrange books.

48. (d)

From statement I, nC1 = 4 => n! / (n-1)!*1! = 4 => n= 4 and nPr = 4

From statement II, nC1 = 1=> n = 1

49. (a)

From statement I, boys= 4 and girls = 3 total ways of arrangement = 7!

From statement II, no. of boys and girls is not given

50. (a)

From statement I, ratio = 45/95 = 9/19

From statement II, first word= TAROT, no. of ways = 5! /2! = 60

Only statement (a) is sufficient to answer the question.

**Click Here For Download permutation & combination Problems with Solutions in pdf**

**Click Here For Download Aptitude Questions on Profit and Loss in pdf**

**Click Here For Download Reverse Syllogism PDF**