# 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

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

While analysing the papers from last many years, I noticed that less than a dozen of questions in every chapter are repeated in every exam.

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
(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
(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?
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. 