Permutation And Combination Questions For IBPS RRB Clerk
Download Top-20 IBPS RRB Clerk Permutation and Combination Questions PDF. Permutation and Combination questions based on asked questions in previous year exam papers very important for the IBPS RRB Assistant exam
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Question 1: What would be the rank of the word ALARM in a dictionary made up of all the possible permutations of the letters of the word?
a) 8
b) 16
c) 55
d) 12
e) None of these
Question 2: If all the possible permutations of the word PAINT are written down in alphabetical order, what is the rank of the word PAINT?
a) 55
b) 49
c) 72
d) 73
e) 61
Question 3: In approximately how many ways can the letters of the word “PERMUTATION” be arranged?
a) 1 crore
b) 2 crores
c) 3 crores
d) 4 crores
e) 5 crores
Question 4: In how many ways can the letters of the word ‘PERMUTATION’ be arranged such that all the consonants are together?
a) $6!/2$
b) $6!^2/2$
c) $5!^2/2$
d) $6!^2$
e) None of the above
Question 5: If all the permutations of the letters of the word “MATHS” are arranged as in a dictionary, what is the rank of the word “MATHS”?
a) 48
b) 53
c) 55
d) 54
e) 50
Question 6: In how many ways can the letters of the word ‘PERMUTATION’ be arranged?
a) 11!/2!
b) 11!
c) 11!*2!
d) 11!/(2!*2!)
e) None of the above
Question 7: A committee of 2 males and 3 female professors has to be formed from a pool of 8 males and 9 female professors. How many different combinations for a committee are possible?
a) 2352
b) 7056
c) 1176
d) 1008
e) None of these
Question 8: In how many ways can a team of 3 people having at least one boy can be formed from a group of 4 boys and 4 girls ?
a) 520
b) 500
c) 510
d) 592
e) 536
Question 9: In how many different ways can the letters of the word ‘AUSTRALIA’ such that all the vowels are together ?
a) 2400
b) 3600
c) 1800
d) 3000
e) 4200
Question 10: In how many ways can a team of 5 people having at least one boy can be formed from a group of 6 boys and 6 girls ?
a) 120
b) 800
c) 720
d) 792
e) 786
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Question 11: What is the probability of selecting 2 red balls and 1 blue ball in the particular given order one at a time from a bag consisting of 6 red balls and 3 blue balls ?
a) 5/26
b) 5/28
c) 5/21
d) 15/28
e) 5/7
Question 12: If a number is randomly selected from the first 50 natural numbers then what is the probability that it is divisible by either 4 or 5 ?
a) 0.33
b) 0.40
c) 0.50
d) 0.25
e) 0.55
Question 13: If all the words that are formed by using the letters in the word “ALIEN” are arranged as listed in the dictionary then how many words are listed before “ILANE” ?
a) 64
b) 60
c) 61
d) 62
e) 63
Question 14: There are 15 chairs in a row and 5 people are to be seated in them such that odd number of chairs should be present between any two people.In how many can it be done ?
a) 70
b) 77
c) 74
d) 72
e) 75
Question 15: There are 4 male and 4 female badminton players and if 2 teams comprising of 1 male and 1 female has to be made then in how many ways can it be done ?
a) 18
b) 80
c) 144
d) 36
e) 72
Question 16: A sum of Rs 80 has to be made by using Rs 1 or Rs 5 coins.In how many different ways can it be done ?
a) 10
b) 17
c) 14
d) 15
e) 16
Question 17: A box contains 5 blue pens, 6 green pens and 10 black pens. The number of ways in which 3 pens can be selected such that at least 2 pens are of the same colour is
a) 1000
b) 1300
c) 1330
d) 1030
e) 1310
Question 18: Find the number of rectangles in the following diagram:
a) 225
b) 315
c) 360
d) 441
e) 482
Question 19: A committee of 5 members is to be formed from 4 men and 4 women. If the committee must have at least 3 men, the number of ways in which the committee can be formed is
a) 24
b) 32
c) 36
d) 40
e) 28
Question 20: Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?
a) 4320
b) 720
c) 2160
d) 120
e) None of these
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Answers & Solutions:
1) Answer (A)
The alphabetical order of the letters is AALMR.
Number of words beginning with AA = 3! = 6
Next would come ALAMR and then ALARM.
Hence, rank would be 8.
2) Answer (D)
All the words starting with A are ahead of PAINT.
The number of such words are 4! = 24
Similarly, all the words starting with I are ahead of PAINT.
The number of such words are 4! = 24
Similarly, all the words starting with N are ahead of PAINT.
The number of such words are 4! = 24
The first word starting with P and using these letters is PAINT.
Hence, its rank is 24+24+24+1 = 73
3) Answer (B)
There are 11 letters in the word PERMUTATION of which T repeats two times. So, the number of ways in which the word can be arranged is 11!/2 = 1,99,58,400
4) Answer (B)
Consonants in the word ‘PERMUTATION’ are P, R, M, T, T, N
Vowels are E, U, A, I, O
Let the consonants be one unit. So, total number of units = 5 + 1 = 6
Number of ways in which these 6 units can be arranged = 6!
Number of ways in which the consonants can be arranged among themselves = 6!/2!
So, total number of ways of arrangement = $6!^2/2$
5) Answer (B)
The alphabetical order of letters is A, H, M, S, T.
Number of words whose first letter is A = 4! = 24
Number of words whose first letter is H = 4! = 24
The next few words will be MAHST, MAHTS, MASHT, MASTH, MATHS.
So, the rank of the word “MATHS” is 24 + 24 + 5 = 53
6) Answer (A)
The total number of letters in the word ‘Permutation’ is 11. Out of these, the letter ‘t’ occurs twice.
So, the number of arrangements of the word = 11!/2!
7) Answer (A)
Number of ways of choosing 2 males from 8 males is $^8C_2$
Number of ways of choosing 3 females from 9 females is $^9C_3$
Total number of ways = $^8C_2 \times ^9C_3 = 2352$
8) Answer (E)
Having at least one boy can be calculated by deleting the case of no boys from the total number of cases i.e
Total number of boys and girls=12
From 12 we have to select 5 ie 12C5=792
No of cases having no boys i.e 6C5=6
So number of cases having at least 1 boy=792-6=786
9) Answer (A)
In AUSTRALIA we have 5 vowels i.e AAAIU and 4 other letters STRL and so assuming all the vowels as single letter we have total of 5 letters to arrange and the internal arrangement of 5 vowels i.e vowels AAAIU can be arranged in 5!/3! Ways
=20 ways
All the 5 letters can be arranged in 5! ways therefore
Total ways=5!*(5!/3!)
=2400 ways
10) Answer (E)
Having at least one boy can be calculated by deleting the case of no boys from the total number of cases i.e
Total number of boys and girls=12
From 12 we have to select 5 ie 12C5=792
No of cases having no boys i.e 6C5=6
So number of cases having at least 1 boy=792-6=786
11) Answer (B)
Given we have 6 red balls and 3 blue balls
So the required probability is $\dfrac{6_{C_{1}}}{9_{C_1}} \times \dfrac{5_{C_{1}}}{8_{C_1}} \times \dfrac{3_{C_{1}}}{7_{C_1}}$
=(6*5*3)/(9*8*7)
=5/28
12) Answer (B)
All the multiples of 4 till 50 are 12 i.e from 4,8…….48
All the multiples of 5 till 50 are 10 i.e from 5,10….50
In the above multiples there are few numbers which are common multiples of 4 and 5 and so they are repeated twice so they are to be removed
LCM of 4 and 5 is 20
Multiples of 20 are 20 and 40 which lie below 50
Therefore total numbers are 12+10-2
=22-2
=20
Probability of selecting a number which is either divisible by 4 or 5 from first 50 natural numbers is 20/50
=2/5
=0.4
13) Answer (C)
Total 5! words can be formed using the 5 letters and so by taking A as the first letter is 4!=24
By taking E as first letter we have 4! words=24
By taking I as first letter and A as the second letter we have 3! Words
By taking I as first letter and E as the second letter we have 3! Words
By taking I as first letter and L as second letter and A as third letter we have 2 words
And the second word is ILANE
So 24+24+6+6+1=61 words come before the word ILANE
14) Answer (B)
In the given condition if all the 5 people are seated in either odd places or even places then the number of chairs between them will be odd and so we have odd places 8 starting from 1,3….15 and even places 7 starting from 2,4…14 and so filing 5 persons in 8 chairs=8C5 =(8*7*6)/(3*2)=56
5 persons sitting in 7 places i.e i 7C5 ways=(7*6)/(1*2)=21
So total number of ways=56+21=77 ways
15) Answer (D)
Two teams are to be selected so 4 players i.e 2 males and 2 females are to be selected.
First select 2 males out of four i.e 4C2=(4*3)/2 =6 ways
then select 2 females out of four i.e 4C2=(4*3)/2 =6 ways
Then first male can be paired with two females and second male can be paired with only one female.
Therefore required ways=6*6*2*1
=72 ways
16) Answer (B)
let the number of Rs 1 coins be x
And the number of Rs 2 coins be y
Therefore x+5y=80
So for every x there should be one y and vice versa
Y take values from y=0 to y=16
I.e for every y from 0 to 16,we will have x
So for y=0 we have x=80
For y=1 we have x=75
So on
Therefore 16+1=17 ways are possible
17) Answer (D)
Number of ways in which at least 2 pens of the same colour can be selected = Total number of ways in which 3 pens can be selected – Number of ways in which 3 pens of different colour can be selected.
There are 5+6+10 = 21 pens in total.
Total number of ways in which 3 pens can be selected = 21C3 = (21*20*19)/(1*2*3) = 1330 ways.
Number of ways in which 3 pens of different colour can be selected = 5*6*10 = 300 ways
Therefore, Number of ways in which at least 2 pens of the same colour can be selected = 1330 – 300 = 1030. Therefore, option D is the right answer.
18) Answer (B)
There are 7 horizontal lines and 6 vertical lines in the diagram.
For a rectangle to be formed, 2 out of these 7 horizontal lines and 2 out of these 6 vertical lines must be selected.
Therefore, total number of rectangles = 7C2*6C2 = 21*15 = 315.
Therefore, option B is the right answer.
19) Answer (E)
The committee can have either 3 men or 4 men.
If 3 men are to be selected, the number of ways in which the committee can be formed is 4C3 * 4C2
= 4*6 = 24 ways.
If 4 men are to be selected, the committee can be formed in 4C4*4C1 = 1*4 = 4 ways.
Therefore, the total number of ways in which the committee can be formed is 24+4 = 28 ways. Hence, option E is the right answer.
20) Answer (A)
Word – DIRECTOR
So “I,E,O” are there are 3! ways to arrange the vowels
Now “D,R,C,T,R” are the remaining alphabets ,
Condition is that the vowels should always be together so we can assume the vowels as a single alphabet/unit say “X” (‘X’=’I,E,O’) so now we have a new word – “D,R,C,T,R,X”
Possible arrangements for this word = 6!
Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of ‘DRCTRX’) $\times$ (Possible arrangements of vowels)
= 6! $\times$ 3! = $720 \times 6 = 4320$
=> Ans – (A)
