BANKERSWAYbankersway.com/Permutation qus .pdf · BANKERSWAY.COM Permutation and Combination Permutation implies arrangement where order of things is important. It includes various

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Permutation and Combination

Permutation implies arrangement where order of things is important. It includes various

patterns like word formation, number formation, circular permutation

etc. Combination means selection where order is not important. It involves selection of

team, forming geometrical figures, distribution of things etc.

Factorial = Factorial are defined for natural numbers, not for negative numbers.

n! = n.(n-1).(n-2) ......... 3.2.1

For example: 1) 4! = 4.3.2.1 = 24

2) 6 ! 4 ! =

6.5.4 ! 4 ! =

6.5 1

= 30

3) 0! = 1

PERMUTATION COMBINATION

Implies Arrangement Implies Selection

Order of things is

important

Order of things is NOT

important

Permutation of three

things a, b and c taking two at

a time are ab, ba, ac, ca,bc

and cb (Order is important).

Combination of three

things a,b and c taking two at

a time are ab, ca and cb

(Order is not important).

nPr = !

( − )!

nCr = !

r ! ( − )!

nPn = n! nCn = 1

N p0 = 1

N C0= 1

Example of Word Formation:

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2

C3 C2

Example - 1: How many new words can be formed with the word "PATNA"?

Solution: In word "PATNA", P,T,N occurs once and A occurs twice.

****Always remember in word formation, if word repeats, number of repetition will be in

denominator.

So, total number of words that can be formed = 5 !

2 ! = 60 i.e.

Therefore, except PATNA there are 59 new words (60-1).

Example – 2 : How many words can be formed from the letters of the word

"EXAMINATION"?

Solution: E, X, M, T, O : Occurs ONCE

A, I, N : Twice

So, total number of words = 11 !

2 ! 2 ! 2 !

(Total number of letters=11 and 3 letters are occurring twice)

Problems for practice

Problem 1: Choose permutation or combination in following terms:-

1) Selection of captain and bowler for a play.

Permutation

2) Selection of four students for a lecture.

Combination

3) Assigning people to their seats during conference.

Permutation

Problem 2: Evaluate 7P2. 4P3

Solution: [ 7 ! ] [ 4 ! ] 5 ! 1 !

⇒(7 × 6). (4 × 3 × 2)

⇒1008

Problem 3: Evaluate 5C2. 3C2

Solution: [ 5 ! ] [ 3 ! ] 3 ! 2 !

⇒ [ 5 × 4 ] . [3] 2 ! 1 !

⇒30

Problem 4: How many ways are there in selecting 5 members from 6 males and 5 females,

consisting 3 males and 2 females?

Solution: This is a case of combination i.e. selecting 3 males from 6 males and 2 females

from 5 females.

⇒Required number of ways = (6 × 5 ) 6 !

3 ! 3 ! × 5 !

2 ! 3 !

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P6

=

⇒ [ 6 × 5 × 4 ] × [ 5 × 4 ] ⇒200

3 × 2 2

Problem 5: How many words can be formed by using letters of the word "DAUGHTER" so

that the vowels come together?

Solution: This is a case of permutation. In a word "DAUGHTER", there are 8 letters

including 3 vowels (AUE)

According to the question, vowels should always come together. Therefore, in this case we

will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words

(one word which is a group of vowels) These 6 words can be arranged in 6 ways

6 6 ! P6

1 ! = 6! = 720 WAYS

Also, three vowels in a group may be arranged in 3! ways

⇒3! = 6 ways

Therefore, required number of words = (720 × 6) = 4320

PROBLEMS WITH SOULATION

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 e) None of these

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


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 e) None of these

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


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

a) 72 b) 144 c) 360


⇒

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


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


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) 1176


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


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


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) 145 e) None of these

12) 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) Data inadequate e) None of these

13) 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


14) A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can

be opened by setting a 4 digit code with the proper combination of each of the 4

rings Maximum how many codes can be formed to open the lock?

a) 49 b) 9p4 c) 94


15) How many different words can be made using the letters of the word

‘HALLUCINATION’ if all consonants are together?

a) 129780 b) 1587600 c) 35600


16) If all S’s come together, then in how many ways the letters of the word

‘SUCCESSFUL be arranged?

a) 10080 b) 40080 c) 2378


17) In how many different ways can 6 different balls be distributed to 4 different boxes,

when each box can hold any number of ball?

a) 2048 b) 1296 c) (24)2


18) What is the total number of 4 digit numbers that can be formed using the digits 0, 1,

2, 3, 4, 5 without repetition, such that the number is divisible by 9?

a) 36 b) 28 c) 15


19) Seven delegates are to address a meeting. If a particular speaker is to speak before

another particular speaker, find the number of ways in which this can be arranged.

a) 1220 b) 2520 c) 3250


20) In how many ways can 15 billiard balls be arranged in a row if 3 are red, 4 are white

and 8 are black?

a) 12 b) 18 c) 96


21) In how many ways can 4 books be arranged out of 16 books on different subjects?

a) 34650 b) 43680 c) 43890


22) Four dice are rolled. The number of possible outcomes in which atleast o ne die

shows 4 is :

a) 671 b) 168 c) 176 d) Data inadequate e) None of these

23) In how many ways can the letters of the word APPLE be arranged?

a) 720 b) 120 c) 60


24) In how many different ways can the letters of the word ‘RUMOUR’ be arranged?

a) 180 b) 90 c) 30


25) A number plate of a vehicle has always a fixed code UP-32 for Lucknow city

followed by the number of particular vehicle which is in two parts. First part is

occupied by 2 English alphabets and second part is occupied by 4 digit numbers

(0001, 0002, …. 9999). If the latest registration number of vehicle [UP-32-SK-

0123] find the number of vehicles registered before this vehicle number in

Lucknow.

a) 2449744 b) 4779644 c) 4669235


26) The number of positive integral solutions of abc = 42 is :

a) 17 b) 27 c) 21

d) 3! × 42 e) None of these

27) A four digit number is formed with the digits 1, 3, 4, 5 without repetition. Find the

chance that the number is divisible by 5 :

a) 3

4

d) 1

16

b) 1

4

e) None of these

c) 9

16

28) 20 girls, among whom are A and B sit down at a round table. The probability that

there are 4 girls between A and B is : a) 17

19

d) 6

19

b) 2

19

e) None of these

c) 13 19

29) The probability that the birthdays of 4 different persons will fall in exactly two

calendar months is : a) 77

1728 b) 17

87 c) 11

144


30) A committee of five persons is to be chosen from a group of 9 people. The

probability that a certain married couple will either serve together or not at all is :

a) 4/9 b) 5/9 c) 13/18


31) In how many different ways can the letters of the word ‘SOFTWARE’ be arranged

in such a way that the vowels always come together?

a) 120 b) 360 c) 1440

d) 13440 e) 720

32) In how many different ways can the letters of the word ‘AUCTION’ be arranged in

such a way that the vowels always come together?

a) 30 b) 48 c) 144


33) In how many ways can 21 books on English and 19 books on Hindu be placed in a

row on a shelf so that two books on Hindi may not be together?

a) 3990 b) 1540 c) 1995


34) Two numbers a and b are chosen at random from the set of first 30 natural numbers.

The probability that 2 - 2 is divisible by 3 is : a) 37

87 b) 47

87 c) 17

29 d) Data inadequate e) None of these

35) From a pack of 52 cards, two are drawn one by one without replacement. Find the

probabilities that both of them are kings. a) 11

21

d) 1

121

b) 13

121

e) None of these

c) 1

221

Solutions

1. Option D

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

x 6 x 5

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

2. Option C

= (525 + 210 + 21)

= 756

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

3. Option D

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

Thus, we have CRPRTN (OOAIO).

= 525 + 7 x 6 x 5

x 6 + 7 x 6

3 x 2 x 1 2 x 1

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

in 5!

= 20 ways. 3!

Required number of ways = (2520 x 20) = 50400

4. Option C

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2)

= 210

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

Each group contains 5 letters.

Number of ways of arranging

5 letters among themselves = 5!

= 5 x 4 x 3 x 2 x 1

= 120.

Required number of ways = (210 x 120) = 25200

5. Option C

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. Option D

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 × 4) + [ 6 × 5

× 4 × 3 ] + [ 6 × 5 × 4 × 4] +[ 6 × 5 ]

2 × 1 2 × 1 3 × 2 × 1 2 × 1

= (24 + 90 + 80 + 15)

= 7 x 6 x 5

x 4 x 3

3 x 2 x 1 2 x 1

[

[ ] [

= 209

7. Option D

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 × 5 × 4) = 20

8. Option C

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

= (8C3 x 10C4)

= 8 × 7 × 6 3 × 2 × 1

× 10 × 9 × 8 × 7 4 × 3 × 2 × 1

= 11760

9. Option C

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 × 6 × 5 2 × 1 +

3 × 2 2 × 1

× 6] + 1

= (45 + 18 + 1)

= 64

10. Option C

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.

]

2 × 1

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

Total number of ways = (6 × 6) = 36

11. Option A

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = [ 7 × 6 × 3] = 63

12. Option C

In the word ‘MATHEMATICS’, we treat the vowels AEAI as oneletter.

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 !

(2!)(2!)

= 10080

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

Number of ways of arranging these letters = 4 !

(2!)

= 12

Required number of words = (10080 × 12) = 120960

13. Option B

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

Required number of ways = (120 × 6) = 720

14. Option C

9 × 9 × 9 × 9 = 94

15. Option B

H L C N T A U I O

L N A I

There are total 13 letters out of which 7 are consonants and 6 are vowels. Also there

are 2L’s, 2N’s, 2A’s and 2I’s

If all the consonants are together then the number of arrangements = 7 !

(2!)(2!)

But the consonants can be arranged themselves in 7 !

(2!)(2!)

Hence the required number of ways = 7 !

ways.

× 7 !

(2!)(2!) (2!)(2!)

16. Option E

S U C E F L

= (1260)2 = 1587600

S U C

S

There are 10 letters in the word SUCCESSFUL and S occurs 3 times, U occurs 2

times and C occurs 2 times.

The letters of the word SUCCESSFUL can be arranged in

= 10 !

3! × 2! × 2! ways

= 151200 ways

17. Option D

Every ball can be distributed in 4 ways. Hence the required number of ways = 4 × 4 × 4 × 4 × 4 × 4

= 46 = 4096

18. Option A

There are two sets of numbers 0, 1, 3, 5 and 0, 2, 3, 4.

Therefore number of 4 digit numbers using digits 0, 1, 3, 5 = 3 × 3 × 2 × 1 = 18 Similarly number of 4 digit numbers using digits 0, 2, 3, 4 = 3 × 3 × 2 × 1 = 18

Hence the total required numbers = 18 + 18 = 36

19. Option B

Total number of ways = 7!

Let A has to speak before B.

Now since there are half of the total cases in which A speaks before B (similarly in

half of the total cases B speaks before A)

Required number of ways = 1

2

× 7! = 2520

20. Option C

Required number of ways = 15 !

3! × 4! × 8!

= 96

21. Option B

16p4 = 43680

22. Option A

Total number of possible outcomes = 64

The number of possible outcomes in which 4 does not appear on any die is 54. Therefore the number of possible outcomes in which atleast one die shows

digit 4 = 64 - 54 = 671

23. Option C

The word APPLE contains 5 letters, 1A, 2P, 1L and 1E.


(1!) (2!) (1!) (1!)

= 60

24. Option A

The word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1U.


(2!) (2!) (1!) (1!)

= 180

25. Option B

Number of vehicles registered upto RZ 9999

= 18 × 26 × (9999)

Number of vehicles registered between SA-0001 and SJ 9999 = 1 × 10 × 9999

Therefore number of vehicles registered before SK-0123

= 18 × 26 × 9999 + 1 × 10 × 9999 + 122

= 9999 × 478 + 122 = 4779644

26. Option B

42 = 2 × 3 × 7

Here each of a, b and c can take 3 values.

Hence the required number of solutions = 3 × 3 × 3 = 27

27. Option B

Total possible number of 4 digits = 4! = 24 The number is divisible by 5 if unit digit itself is 5. Therefore we fix 5 at unit place

and then remaining 3 places can be filled up in 3! Ways.

Hence, the required probability = 3!

4! =

6 =

1 24 4

28. Option B

20 girls can be seated around a round table in 19! Ways.

So, exhaustive number of cases = 19!

Excluding A and B, out of remaining 18 girls, 4 girls can be selected 18C4 ways

which can be arranged in 4! Ways. Remaining 20 - (4 - 2) = 14 girls can be arranged

in 14! Ways. Also A and B mutually can be arranged in 2! Ways. Required number of arrangements = 18C4 × 4! × 2! × 14!

= 18! × 2

Required probability = 18! × 2

= 2

19! 19

29. Option A

Since a person’s birthday can fall in any of the 12 months.

So, total number of ways = 124

Now, any two months can be chosen in 12C2 ways. The 4 persons birthday can fall in

these two months in 24 ways. Out of these 24 ways there are two ways when all of the four birthdays fall in one month. So, favourable number of ways = 12C2 × (24 - 2)

= 77

1728

30. Option A

124

Total number of ways in which 5 people can be chosen out of 9 people = 9C5 = 126

Number of ways in which the couple serves the committee = 7C3 × 2C2 = 35

Number of ways in which the couple does not serve the committee = 7C5 = 21 Favourable number of cases = 35 + 21 = 56

Hence, the required probability = 56

= 4

126 9

31. Option E

The word ‘SOFTWARE’ contains 8 different letters.

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

Thus, we have to arrange the letters SFTWR (OAE).

Now, 5 letters can be arranged in 6! = 720 ways The vowels (OAE) can be arranged among themselves in 3! = 6 ways.


32. Option D

The word AUCTION has 7 different letters. When the vowels AUIO are always together, they can be supposed to form one

letter.

Then,we have to arrange the letters CTN (AUIO).

Now, 4 letters can be arranged in 4! = 24 ways.

The vowels (AUIO) can be arranged among themselves in 4! = 24 ways.


33. Option B

In order that two books on Hindi are never together, we must place all these books

as under:

X E X E X E X …. X E X

Where E denotes the position of an English book and X that of a Hindi book.

Since there are 21 books on English, the number of places marked X are therefore, 22

Now, 19 places out of 22 can be chosen in 22C19 = 22C3 = 22 × 21 × 20

= 1540 ways 3 × 2 × 1

Hence, the required number of ways = 1540

34. Option B

Out of 30 numbers 2 numbers can be chosen in 30C2 ways.

So, exhaustive number of cases = 30C2 = 435

Since 2 - 2 is divisible by 3 if either a and b are divisible by 3 or none of a and b

is divisible by 3. Thus, the favourable numbers, of cases = 10C2 + 20C2 = 235 Hence, required probability =

235 =

47

35. Option C

435 87

Required probability = 4

× 3

= 1

52 51 221

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BANKERSWAYbankersway.com/Permutation qus .pdf · BANKERSWAY.COM Permutation and Combination Permutation implies arrangement where order of things is important. It includes various