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There are 2 ways from A to B and 3 ways from B to C. In how many ways can a person reach from A to C? Total no of ways is = 2 * 3 = 6 Why multiplication? The events are independent A C If a coin is tossed 3 times how many results are possible? H T H T T H H T T H H T H T B Total no of results = 8 1

Permutations & Combinations

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Page 1: Permutations & Combinations

There are 2 ways from A to B and 3 ways from B to C. In how many ways can a person reach from A to C?

Total no of ways is = 2 * 3 = 6 Why multiplication?The events areindependent

A C

If a coin is tossed 3 times how many results are possible?

H

T

H

T

T

H

H

T

TH

H

TH

T

B

Total no of results = 8

1

Page 2: Permutations & Combinations

A die is rolled 3 times. Number of outcomes possible is -----(a)1 (b) 3(c) 18 (d) 216

No of results is 6*6*6 = 216

In how many different ways can 10 different varieties of chocolates be distributed to 2 friends?(a) 45 (b) 90 (c) 19(d) 1024No. of ways = 2*2*………. 10 times = 1024

A die is rolled 4 times. Find the no. of distinct sums possible on the top faces.(a)24 (b)1(c) 21 (d) 1296Minimum sum = 4Maximum sum = 24 Number of distinct sums possible= 21.

2

A coin is rolled n times. Number of outcomes is 2n

A die is rolled n times. Number of outcomes is 6n

Page 3: Permutations & Combinations

In how many ways can ABC seat themselves in 3 places. ABC BCA CAB ACB BAC CBA No. of ways = 6

= 3*2*1=6

In how many ways can ABC seat themselves in 3 places, if A& B are identical Red balls and c is a green ball ? RRG RGR GRR RGR RRG GRR No. of ways = 3

Using 1,2,3 how many 3 digit numbers can be formed?Two answers possible.Case 1: With repetition Total = 3*3*3=27Case 2: Without repetition.

Total =3*2*1=6

3

How many ways can 3 people be seated in 4 chairs?No. of ways = 4*3*2 =24.

How many ways can 4 people be seated in 3 chairs?No. of ways = 4*3*2 =24.

Page 4: Permutations & Combinations

4

Permutations

Items non distinctItems distinct

With repetition

Without repetition

With repetition

Without repetition

Page 5: Permutations & Combinations

In how many ways can ABCD dance one after another?No of ways = 4*3*2*1 = 24

5

ABCD BACD CABD DABC

ABDC BADC CADB DACB

ACBD BCAD CBAD DBAC

ACDB BCAB CBDA DBCA

ADBC BDAC CDAB DCAB

ADCB BDCA CDBA DCBA

In how many ways can ABCD be arranged such that A&B are always together?Consider AB as one set. No of ways of arranging becomes = 3!*2! =12

In how many ways can ABCD dance one after another such that A dances before B?

No of ways = 24/2 = 12

Why divide by 2?

In how many ways can ABCD dance one after another such that A dances before B and B dances before C?

No of ways = 24/3! = 4

Why divide by 6?

Page 6: Permutations & Combinations

A question paper has 4 questions and each question has 2 choices. In how many a ways can a person answer one or more question?

6

No of questions answered

Selection TOTAL

0 4c0 1 4c0*20

1 4c1 21 4c1*21

2 4c2 22 4c2*22

3 4c3 23 4c3*23

4 4c4 24 4c4*24

First question cane be answered in 3 ways namely 1a or 1b or 1 not answered. Similarly 2nd , 3rd & 4th questions can be answered in 3 ways each.As answering each question is an independent event, total number is 34 – 1 = 80 .

Why subtract 1?

4c1*2 + 4c2*22 + 4c3*23 + 4c4*24

= 8+24+32+16 = 80 ways

These 9 ways including one case where no question is answered. Hence 32 – 1 = 8 .

1a 2a

1a 2b

1a 2 na

1b 2a

1b 2b

1b 2 na

1na 2a

1na 2b

1na 2 na

Page 7: Permutations & Combinations

A question paper has 60 questions and each question has 4 choices. In how many a ways can a person answer two or more question?Each question can be answered in 5 ways. Total number of ways of attempting the paper is 560

Not attempting any question is 1 way.Attempting one question is 60* 4 = 240.Total number of ways = 560 – 1 – 240 = 560 – 241

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A pool of 10 cops are available for providing security to a VIP. At least 2 cop are needed to provide a security. In how many ways can the security be provided to the VIP?Total number of ways of deploying the cops is 210 , including one way where no cop is deployed. Alotting one cop can be provided in 10 ways.Total number of ways = 210 – 1 – 10 = 210 – 11

Page 8: Permutations & Combinations

There are 3 apples and 4 bananas . In how many ways can I pick 1 or more fruits? (a) 12 (b) 128 (c) 19 (d) 20

No. of ways = (3+1)(4+1)-1 = 19.

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Anil wrote down all the possible three-digit numbers with distinct digits on a black board. Of these numbers, Biswas erased all the numbers whose first and last digits were either both even or both odd. How many numbers were left on the board? (1)  450 (2)  360 (3)  400 (4)  320 (5)  540First Digit Odd : 5*8*5 = 200First Digit even : 4*8*5 = 160Total = 360

Page 9: Permutations & Combinations

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How many 4-digit odd numbers are possible such that hundred’s digit is 2 more than ten’s digit?1) 315 2) 340 3) 350 4) 400 5 )360The units digit can be filled in 5 ways.The hundred’s & tens digits can be filled in 8 ways.The thousands digit can be filled in 9 ways.Total numbers is 9 *8*1* 5 = 360

1000’s

100’S

10’S

Units

1 23456789

2 01

3

5

7

9

3 1

4 2

5 3

6 4

7 5

8 6

9 7

Page 10: Permutations & Combinations

10

Over a period of 11 days, a vegetable vendor visited exactly 10 villages for selling vegetables. On each day he visited exactly one village, but he did not revisit any village within two days of visiting it. In how many ways could he have visited the villages in the period of 11 days?(1) 10(9)10 (2) 10C2(8)9 (3) 10(8)10

(4) 720(7)8 (5) 90(8)9

10*9*8*8*8*8*8*8*8*8*8 = 90(8)9

In the grid shown alongside, six X's have to be placed such that each row contains at least one X. In how many ways can this be done? (1) 180 (2) 176 (3) 226 (4) 352 (5) 188

Six x’s can be placed in 10 cells in 10c6

Let us identify how a row can be

kept empty and subtract from

total arrangements.

Row 1 free. No of arrangement : 1

Row 3 free. No of arrangement : 1

Row 2 free. No of arrangement : 8c6

210 – 1 - 1 – 28 = 180

Page 11: Permutations & Combinations

Among all the four-digit natural numbers divisible by 24, how many have the number 24 in them?(1) 24 (2) 26 (3) 28 (4) 25 (5) None of these

11

Th H T U

2 4 0 0

2 4 2 4

2 4 4 8

2 4 7 2

2 4 9 6

Th H T U

1 2 2 4

1 8 2 4

2 4 2 4

3 0 2 4

3 6 2 4

4 2 2 4

4 8 2 4

5 4 2 4

6 0 2 4

6 6 2 4

7 2 2 4

3 2 4 0

6 2 4 0

9 2 4 0

1 2 4 8

4 2 4 8

7 2 4 8

Th H T U

7 8 2 4

8 4 2 4

9 0 2 4

9 6 2 4

There are 25 values.

Page 12: Permutations & Combinations

The circumference of a circle is divided into 26 equal parts by marking 26 equidistant points on it. Now, using these points as vertices, triangles are drawn suchthat the circumcentre of each of those triangles lies on one of its sides. How many such triangles can be drawn?(1)156 (2)676 (3)182 (4)650 (5)312

12

For each diameter there will be 24 such triangles. No of diameters = 13. Totally there will be 512 triangles.

Page 13: Permutations & Combinations

How many numbers can be formed <5556 using the digits 1 to 5?

13

How many four digit numbers > 4567 can be formed using the digits 3,4,5,6,7 not more than once?(a) 72 (b) 82 (c) 84 (d) 86.

1000’s

100’s

10’s

units

4 digitNos.

5 5 5 5 625

3 digitNos.

5 5 5 125

2 digitNos.

5 5 25

1digitNos.

5 5

Total no of numbers 780

4 Digit Nos

1000’s

100’s

10’s

units 86

Starting with 5 4 3 2 24

Starting with 6 4 3 2 24

Starting with 7 4 3 2 24

4573& 4576 2

4635,4637,4653,4657, 4673,4675

6

4635,4637,4653, 4657,4673,4675

6

Page 14: Permutations & Combinations

How many ways can we give 15 distinct books to 3 students?No. of ways = 15C5* 10C5

= 15!/(5!) (5!) (5!)

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How many ways can we make 3 equal parcels with 15 distinct books?No. of ways = 15C5* 10C5 / 3!

= 15!/(5!) (5!) (5!) (3!)

How many ways the letters of the word GAIN be permuted so that vowels are together? No of words = 3! * 2!

= 12How many ways the letters of the word ALTER be permuted so that vowels are together? No of words = 4! * 2!

= 48How many ways the letters of the word alert be permuted so that consonants are together? No of words = 3! * 3!

= 36

Page 15: Permutations & Combinations

How many ways the letters of the word avert be permuted so that vowels are not together?Total No of words = 5!No of words with vowels being together = 48No of words with vowels being together = 72Aliter There are 3 consonants and 2 vowels.

Vowels can occupy places marked “ ” in 12 . Consonants can be filled in boxes in 6 ways. Totally in 72 ways.

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How many ways the letters of the word DAUGHTER be permuted so that no two vowels are together? There are 5 consonants and 3 vowels.

Vowels can occupy places marked “ ” in 6*5*4.

Consonants can be filled in boxes in 5! ways. Totally in 14400 ways.

Page 16: Permutations & Combinations

How many nos. >999 and < 4000 can be formed using the digits 0,1,2,3,4 if repetition of digits is allowed?(a) 499 (b) 500 (c ) 375 (d) 376No of numbers ::: 3*5*5*5 = 375

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Find the rank of the word COCHIN as arranged in a dictionary?No. of words that begin with C = 5! = 120No. of words that begin with CC=4! = 24No. of words that begin with CH=4! = 24No. of words that begin with CI = 4!= 24No of words that begin with CN= 4! = 24The next word is COCHINRank of COCHIN is 217.

In how many ways 5 Maths books & 2 physics books can be arranged in a book shelf such that physics books are not together? There are 5 Maths books and 2 Physics books.

Physics books can occupy places marked “ ” in 6*5.Consonants can be filled in boxes in 5! ways. Totally in 3600 ways.

Page 17: Permutations & Combinations

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In how many ways 4 Maths books & 3 physics books can be arranged in a book shelf such that physics books are not together? There are 4 Maths books and 3 Physics books.

Physics books can occupy places marked “ ” in 5*4*3.Consonants can be filled in boxes in 4! ways. Totally in 1440 ways. In how many ways 4 boys & 4 girls sit together such that no two boys & no two girls together?

Boys can sit in places marked in “ ” in 4*3*2*1.Girls can sit in places marked in “ ” in 4*3*2*1.The arrangement can start with a boy or a girl.Therefore totally in 2! 4! 4! = 1152

Page 18: Permutations & Combinations

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A five digit number divisible by 3 is to be formed using the numbers 0 , 1 , 2 , 3 , 4 , 5 without repetition . Find the number of such numbers?For a number to be divisible by 3 sum of the digits should be divisible by 3.Minimum sum possible = 10.Maximum sum possible = 15.So expected sums are 12 and 15.Sum of 12 is possible when the digits chosen are 0,1,2,4,5.Number of 5 digit numbers using the above digits are 4 x 4 ! = 96.Sum of 15 is possible if the digits chosen are 1,2,3,4,5.Number of 5 digit numbers using these digits are 5 ! = 120.Hence total number of 5 digit numbers div. by 3 using the given digits are = 216.

Page 19: Permutations & Combinations

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Find the sum of all 4 digit numbers formed using the digits 1, 2, 3& 4 without repetition.Sum of the digits in the units place = 3! (1+2+3+4)* 1Sum of the digits in tens place = 3! (1+2+3+4)*10Sum of the digits in hundreds place = 3! (1+2+3+4)*100Sum of the digits in th.place = 3! (1+2+3+4)*1000Sum of all the nos. = 3! (1+2+3+4)(1111) = 66660.

Find the sum of all 4 digit numbers formed using the digits 1, 2, 3& 4 with repetition.

Sum of the digits in the units place = 43 (1+2+3+4)* 1

Sum of the digits in tens place = 43 (1+2+3+4)* 10

Sum of the digits in hundreds place = 43 (1+2+3+4)* 100

Sum of the digits in th.place = 43 (1+2+3+4)*1000

Sum of all the nos. = 43 (1+2+3+4)*(1111) = 711040

Page 20: Permutations & Combinations

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Find the no. of words using the letters of the word VOWELS so that no 2 consonants are together? No. of ways = 0.

Find the no. ways 32 boys can be seated so that selected 16 are in the I row & the rest in II row?No. of arrangements = 16!*16!

Find the no. of ways 32 boys can be seated in 2 rows? No. of arrangements = 32!

How many ways can 3 different prizes can be given to 4 boys?

The I prize can be given to any of the 4 boys.The II prize can be given to any of the 4 boys.The III prize can be given to any of the 4 boys.No. of ways prizes can be distributed = 43

Page 21: Permutations & Combinations

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How many ways can 8 different letters be posted in 6 psot boxes?

The 1 letter can be posted in 6 ways. Similarly other letters can be posted in 6 ways.

No. of ways letters can be posted = 68

How many ways can 3 different prizes can be given to 4 boys, if each prize has 3 positions?

The I prize 1st position can be given to any of the 4 boys.The I prize 2nd position can be given to any of the 3 boys.The I prize 3rd position can be given to any of the 2 boys.The 1 prize can be given in 24 ways.The 2 prize can be given in 24 ways.The 3 prize can be given in 24 ways.No. of ways prizes can be distributed = 243

Page 22: Permutations & Combinations

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Find the number of odd proper divisors of 3 m x 6 n x 21 p .

3m x6n x 21p = 3m x 2n x 3n x 3p x 7p = 2n x 3 m+n+p x 7p

Hence number of odd proper divisors = number of ways of selecting any number of 3’s and 7’s from ( m+n+p) identical 3’s and p identical 7’s. = ( m+n+p+1)(p+1) – 1.

Page 23: Permutations & Combinations

Find the number of ways 10 people can be seated around a table where the chairs are numbered so that 2 particular persons are not seated together?

When chairs are numbered circular permutation is the

same as linear permutation.

Hence the number of arrangements are = 10!

The number of arrangements where 2 particular

persons are together = 9 ! X 2 !

But if the 2 selected persons occupy end positions

when it is circular arrangement they are together.

Hence we need to remove those cases as well.

Number of ways they can occupy end positions = 2! X

8!.

Hence required arrangements = 10! – 9! X 2! – 2! X 8!.

Page 24: Permutations & Combinations

Find the maximum number points of intersection of 4 circles and 4 straight lines? Four lines can intersect each other at 4 C 2 = 6 points.

Four Circles can intersect each other at 4 C 2 x 2 = 12 points.

A line can cut 4 circles at a max. of 8 points.4 lines will cut 4 circles at 4 x 8 =32 points.Hence maximum number of points of intersection can be = 6 + 12 +32 = 50.

• ••

•••

• •

Page 25: Permutations & Combinations

Consider the equation a x 2 + b x + 1 = 0 .If a and b are both randomly chosen real numbers between o and 4, what is the probability that the equation has no real roots? The condition for roots to be non real is b 2 < 4a.Hence the required probability is = Shaded area / Area of the square o x

y

Y2 =4x

(4,0)

(0,4)

32/3=

=∫2 √X DX0

44/3[X3/2 ]

04

= [4/3][8 ]

Area of Shaded portion

Required probability

= 32/[3*16]

= 2/3

Page 26: Permutations & Combinations

27).Find the no. of ways 4 boys & 4 girls can be seated around a table so that no two boys are together?

B

B

B

B

G

G

G

GG

No of arrangements = 3!*4!

26

Page 27: Permutations & Combinations

28). 29).If 10 couples are invited for a party so that either wife alone attends or couple attend the party.How many ways the party can be arranged?

(a) 1024 (b) 29 (c) 19 (d) 30 (e)none of the above No.of different parties can be arranged = 310-1.

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