## Elitmus Previous Years Questions on Permutations and Combination

# Permutations And Combinations 01

##
01.Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between the greatest and smallest numbers that can be formed?
(a) 19800
(b) 41976
(c) 32976
(d) None of these

02.In how many ways can Eight Directors, Vice-Chairman and Chairman of a firm be seated at a round table, if the Chairman has to sit between the Vice-Chairman and a Director?
(a) 9! x 2
(b) 2 x 8!
(c) 2 x 7!
(d) None of these

03. A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be 3 exactly girls in the invitees?
(a) 320
(b) 160
(c) 80
(d) 200
04.
Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?
(a) 8
(b) 10
(c) 15
(d) 22
05.
A, B, C, D are four towns, any three of which are non-collinear. Then, the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is?
(a) 7
(b) 8
(c) 9
(d) 24
06.
If a 4-digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?
(a) 1/12
(b) 1/4
(c) 1/6
(d) None of these

07. A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
(a) 6666600
(b) 6666660
(c) 6666666
(d) None of these

08.139 persons have signed for an elimination tournament. All players are to be paired up for the first round, but because 139 is an odd number one player gets a bye, which promotes him to the second round, without actually playing in the first round. The pairing continues on the next round, with a bye to any player left over. If the schedule is planned so that a minimum number of matches is required to determine the champion, the number of matches which must be played is
(a) 136
(b) 137
(c) 138
(d) 139

09.A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of different size. The probability that the red ball selected is the smallest red ball?
(a) 1/18
(b) 1/3
(c) 1/6
(d) 2/3

10. A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?
(a) 3
(b) 4
(c) 5
(d) 6

Answers :
1. C
2. B
3. B
4. D
5. D
6. A
7. A
8. C
9. C
10. D

###
Permutation And Combinations 02

01. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with the every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is?
(a) 200
(b) 216
(c) 235
(d) 256

02.
A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is?
(a) 12 x 81
(b) 16 x 192
(c) 20 x 125
(d) 24 x 216
03.
An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …. 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion, when read upside down- for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?
(a) 80
(b) 78
(c) 71
(d) 69
04.
How many numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?
(a) 496
(b) 486
(c) 1084
(d) 728
05.
In how many ways is it possible to choose a white square and a black square on a chess-board so that the squares must not lie in the same row or column?
(a) 56
(b) 896
(c) 60
(d) 768
06.
How many four-letter computer passwords can be formed using only the symmetric letters. (no repetition allowed) (Symmetric letters :- A, H, I, M, O, T, U, V, W, X AND Z)
(a) 7920
(b) 330
(c) 14640
(d) 419430
07.
How many three-letter computer password can be formed with at least one symmetric letter ? (Symmetric letters: A, H, I, M, O, T, U, V, W, X and Z)
(a) 990
(b) 2730
(c) 12870
(d) 1560000
08.
For a scholarship, at the most n candidates out of 2n+1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is ?
(a) 3
(b) 4
(c) 6
(d) 5
09.
Ten points are marked on a straight line and 11 points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?
(a) 495
(b) 550
(c) 1045
(d) 2475
10.
How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at unit’s place must be greater than the in the ten’s place?
(a) 54
(b) 60
(c) 17
(d) 2 x 4!
Answers :Find Detailed Solutions at the end of the page
1. A
2. A
3. D
4. D
5. D
6. A
7. C
8. A
9. C
10. B

Set 1 Detailed Solutions

1.

Greatest five digit number : 43210

Smallest five digit number : 10234

Difference = 43210-10234= 32976

(c)

2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!

(b)

3. 3 Girls can be selected out of 5 girls in 5C3 ways.

Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.

Hence required number of ways is = 5C3 x (2)4 = 160

(b)

4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.

Two adjacent boxes with blue can be got in 4 ways.

i.e. (1, 2) (2, 3) (3, 4) (4, 5)

Three adjacent boxes with blue can be got in 3 ways

i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)

Four adjacent boxes with blue can be be got in 2 ways.

i.e. (1234) , (2345)

and five boxes with blue can be got in 1 way.

Hence total number of ways of filling the boxes such that adjacent boxes have blue

(4 + 3 + 2 + 1)=10

(d)

5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.

(d)

6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.

Hence required probability = 2!/4! = 1/12

(a)

7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers

=600(1+10+100+1000+10000)=6666600

(a)

8. Required number of matches played will be (139 – 1)=138

(c)

9. Required probability is = 1/6

(c)

10. Let the number of students in the front row be x and the number of rows be n.

Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.

Now we have to check for each value of n=3, 4, 5, 6

Firstly take n=3

x + (x – 3) + (x – 6) =630

» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)

Likewise if n=4

x + (x – 3) + (x – 6)+(x – 9) =630

» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)

Likewise if n=5

x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630

» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)

Likewise if n=6

x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630

» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)

(d)

Set 2 Detailed Solutions

1. Let there be m boys and n girls.

Then nC2 = 45 » n(n – 1)=90» n =10

mC2 = 190 » m(m – 1) = 380 » m=20

Number of games played between one boy and one girl

= 10C1 x 20C1= 10 x 20=200

(a)

2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).

Similarly, there are 3 ways to colour each of the remaining stripes.

» The number of ways the flag can be coloured is

4 x (3)5 = (12)(3)4

(a)

3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.

Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.

Total number of ways confusion can arise = 4 x 3=12

Thus, required answer= 81-12=69

(d)

4.

Number of ways for selecting single digit=2

Number of ways for selecting two digit= 2 x 3=6

Number of ways for selecting three digit= 2 x 3 x 3=18

Number of ways for selecting four digit= 2 x 3 x 3 x 3=54

Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162

Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486

Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728

(d)

5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one white and one black square on the chess.

32C1 x 32C1= 32 x 32=1024

Number of ways in which square lies in the same row

White square=4

Black square=4

Number of rows=8

4C1 x 4C1 x 8 = 128

»Number of ways in which square lies in the same column= 128

Total number in which square lie on the same row or same column= 128 + 128=256.

(d)

6. Ist place of the four letter password can be filled in 11 ways.

IInd place of four letter password can be filled in 10 ways.

IIIrd place of four letter password can be filled in 9 ways.

IVth place of four letter password can be filled in 8 ways.

Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways

(a)

7. Three letter password from 26 letters can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.

Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.

(c)

8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.

» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5

Since n cannot be a fraction. Hence n=3.

(a)

9. Required number of triangles formed

10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045

(c)

10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.

Hence required number of ways = 4!

However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.

Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers

= 4! + (3) x 3! + (2)3! + 3!

= 4! + 6 x 3! = 24 + (6 x 6) = 60

(b)

01.Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between the greatest and smallest numbers that can be formed?

(a) 19800

(b) 41976

(c) 32976

(d) None of these

02.In how many ways can Eight Directors, Vice-Chairman and Chairman of a firm be seated at a round table, if the Chairman has to sit between the Vice-Chairman and a Director?

(a) 9! x 2

(b) 2 x 8!

(c) 2 x 7!

(d) None of these

03. A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be 3 exactly girls in the invitees?

(a) 320

(b) 160

(c) 80

(d) 200

04.

Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?

(a) 8

(b) 10

(c) 15

(d) 22

05.

A, B, C, D are four towns, any three of which are non-collinear. Then, the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is?

(a) 7

(b) 8

(c) 9

(d) 24

06.

If a 4-digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

(a) 1/12

(b) 1/4

(c) 1/6

(d) None of these

07. A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

(a) 6666600

(b) 6666660

(c) 6666666

(d) None of these

08.139 persons have signed for an elimination tournament. All players are to be paired up for the first round, but because 139 is an odd number one player gets a bye, which promotes him to the second round, without actually playing in the first round. The pairing continues on the next round, with a bye to any player left over. If the schedule is planned so that a minimum number of matches is required to determine the champion, the number of matches which must be played is

(a) 136

(b) 137

(c) 138

(d) 139

09.A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of different size. The probability that the red ball selected is the smallest red ball?

(a) 1/18

(b) 1/3

(c) 1/6

(d) 2/3

10. A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

(a) 3

(b) 4

(c) 5

(d) 6

Answers :

1. C

2. B

3. B

4. D

5. D

6. A

7. A

8. C

9. C

10. D

### Permutation And Combinations 02

01. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with the every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is?

(a) 200

(b) 216

(c) 235

(d) 256

02.

A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is?

(a) 12 x 81

(b) 16 x 192

(c) 20 x 125

(d) 24 x 216

03.

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …. 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion, when read upside down- for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

(a) 80

(b) 78

(c) 71

(d) 69

04.

How many numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?

(a) 496

(b) 486

(c) 1084

(d) 728

05.

In how many ways is it possible to choose a white square and a black square on a chess-board so that the squares must not lie in the same row or column?

(a) 56

(b) 896

(c) 60

(d) 768

06.

How many four-letter computer passwords can be formed using only the symmetric letters. (no repetition allowed) (Symmetric letters :- A, H, I, M, O, T, U, V, W, X AND Z)

(a) 7920

(b) 330

(c) 14640

(d) 419430

07.

How many three-letter computer password can be formed with at least one symmetric letter ? (Symmetric letters: A, H, I, M, O, T, U, V, W, X and Z)

(a) 990

(b) 2730

(c) 12870

(d) 1560000

08.

For a scholarship, at the most n candidates out of 2n+1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is ?

(a) 3

(b) 4

(c) 6

(d) 5

09.

Ten points are marked on a straight line and 11 points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?

(a) 495

(b) 550

(c) 1045

(d) 2475

10.

How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at unit’s place must be greater than the in the ten’s place?

(a) 54

(b) 60

(c) 17

(d) 2 x 4!

Answers :Find Detailed Solutions at the end of the page

1. A

2. A

3. D

4. D

5. D

6. A

7. C

8. A

9. C

10. B

Set 1 Detailed Solutions

1.

Greatest five digit number : 43210

Smallest five digit number : 10234

Difference = 43210-10234= 32976

(c)

2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!

(b)

3. 3 Girls can be selected out of 5 girls in 5C3 ways.

Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.

Hence required number of ways is = 5C3 x (2)4 = 160

(b)

4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.

Two adjacent boxes with blue can be got in 4 ways.

i.e. (1, 2) (2, 3) (3, 4) (4, 5)

Three adjacent boxes with blue can be got in 3 ways

i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)

Four adjacent boxes with blue can be be got in 2 ways.

i.e. (1234) , (2345)

and five boxes with blue can be got in 1 way.

Hence total number of ways of filling the boxes such that adjacent boxes have blue

(4 + 3 + 2 + 1)=10

(d)

5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.

(d)

6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.

Hence required probability = 2!/4! = 1/12

(a)

7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers

=600(1+10+100+1000+10000)=6666600

(a)

8. Required number of matches played will be (139 – 1)=138

(c)

9. Required probability is = 1/6

(c)

10. Let the number of students in the front row be x and the number of rows be n.

Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.

Now we have to check for each value of n=3, 4, 5, 6

Firstly take n=3

x + (x – 3) + (x – 6) =630

» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)

Likewise if n=4

x + (x – 3) + (x – 6)+(x – 9) =630

» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)

Likewise if n=5

x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630

» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)

Likewise if n=6

x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630

» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)

(d)

Set 2 Detailed Solutions

1. Let there be m boys and n girls.

Then nC2 = 45 » n(n – 1)=90» n =10

mC2 = 190 » m(m – 1) = 380 » m=20

Number of games played between one boy and one girl

= 10C1 x 20C1= 10 x 20=200

(a)

2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).

Similarly, there are 3 ways to colour each of the remaining stripes.

» The number of ways the flag can be coloured is

4 x (3)5 = (12)(3)4

(a)

3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.

Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.

Total number of ways confusion can arise = 4 x 3=12

Thus, required answer= 81-12=69

(d)

4.

Number of ways for selecting single digit=2

Number of ways for selecting two digit= 2 x 3=6

Number of ways for selecting three digit= 2 x 3 x 3=18

Number of ways for selecting four digit= 2 x 3 x 3 x 3=54

Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162

Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486

Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728

(d)

5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one white and one black square on the chess.

32C1 x 32C1= 32 x 32=1024

Number of ways in which square lies in the same row

White square=4

Black square=4

Number of rows=8

4C1 x 4C1 x 8 = 128

»Number of ways in which square lies in the same column= 128

Total number in which square lie on the same row or same column= 128 + 128=256.

(d)

6. Ist place of the four letter password can be filled in 11 ways.

IInd place of four letter password can be filled in 10 ways.

IIIrd place of four letter password can be filled in 9 ways.

IVth place of four letter password can be filled in 8 ways.

Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways

(a)

7. Three letter password from 26 letters can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.

Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.

(c)

8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.

» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5

Since n cannot be a fraction. Hence n=3.

(a)

9. Required number of triangles formed

10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045

(c)

10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.

Hence required number of ways = 4!

However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.

Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers

= 4! + (3) x 3! + (2)3! + 3!

= 4! + 6 x 3! = 24 + (6 x 6) = 60

(b)

Set 1 Detailed Solutions

1.

Greatest five digit number : 43210

Smallest five digit number : 10234

Difference = 43210-10234= 32976

(c)

2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!

(b)

3. 3 Girls can be selected out of 5 girls in 5C3 ways.

Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.

Hence required number of ways is = 5C3 x (2)4 = 160

(b)

4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.

Two adjacent boxes with blue can be got in 4 ways.

i.e. (1, 2) (2, 3) (3, 4) (4, 5)

Three adjacent boxes with blue can be got in 3 ways

i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)

Four adjacent boxes with blue can be be got in 2 ways.

i.e. (1234) , (2345)

and five boxes with blue can be got in 1 way.

Hence total number of ways of filling the boxes such that adjacent boxes have blue

(4 + 3 + 2 + 1)=10

(d)

5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.

(d)

6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.

Hence required probability = 2!/4! = 1/12

(a)

7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers

=600(1+10+100+1000+10000)=6666600

(a)

8. Required number of matches played will be (139 – 1)=138

(c)

9. Required probability is = 1/6

(c)

10. Let the number of students in the front row be x and the number of rows be n.

Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.

Now we have to check for each value of n=3, 4, 5, 6

Firstly take n=3

x + (x – 3) + (x – 6) =630

» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)

Likewise if n=4

x + (x – 3) + (x – 6)+(x – 9) =630

» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)

Likewise if n=5

x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630

» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)

Likewise if n=6

x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630

» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)

(d)

Set 2 Detailed Solutions

1. Let there be m boys and n girls.

Then nC2 = 45 » n(n – 1)=90» n =10

mC2 = 190 » m(m – 1) = 380 » m=20

Number of games played between one boy and one girl

= 10C1 x 20C1= 10 x 20=200

(a)

2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).

Similarly, there are 3 ways to colour each of the remaining stripes.

» The number of ways the flag can be coloured is

4 x (3)5 = (12)(3)4

(a)

3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.

Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.

Total number of ways confusion can arise = 4 x 3=12

Thus, required answer= 81-12=69

(d)

4.

Number of ways for selecting single digit=2

Number of ways for selecting two digit= 2 x 3=6

Number of ways for selecting three digit= 2 x 3 x 3=18

Number of ways for selecting four digit= 2 x 3 x 3 x 3=54

Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162

Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486

Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728

(d)

5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one white and one black square on the chess.

32C1 x 32C1= 32 x 32=1024

Number of ways in which square lies in the same row

White square=4

Black square=4

Number of rows=8

4C1 x 4C1 x 8 = 128

»Number of ways in which square lies in the same column= 128

Total number in which square lie on the same row or same column= 128 + 128=256.

(d)

6. Ist place of the four letter password can be filled in 11 ways.

IInd place of four letter password can be filled in 10 ways.

IIIrd place of four letter password can be filled in 9 ways.

IVth place of four letter password can be filled in 8 ways.

Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways

(a)

7. Three letter password from 26 letters can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.

Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.

(c)

8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.

» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5

Since n cannot be a fraction. Hence n=3.

(a)

9. Required number of triangles formed

10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045

(c)

10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.

Hence required number of ways = 4!

However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.

Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers

= 4! + (3) x 3! + (2)3! + 3!

= 4! + 6 x 3! = 24 + (6 x 6) = 60

(b)

Related Pages:

## No comments:

## Post a Comment