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In how many ways can 7 different balls be distributed in 5 different boxes if box 3 and box 5 can contain only one and two number of balls respectively and rest of the boxes can contain any number of balls?

A.

10100

B.

6200

C.

8505

D.

12800

Answer with explanation

Answer: Option CExplanation

One ball for box 3 can be selected in ^{7}C_{1} ways.

Two balls for box 5 can be selected in ^{6}C_{2} ways.

Remaining balls = 4

Remaining boxes = 3

In these 4 balls, 1^{st} ball can be put in any of these 3 boxes.

Similarly 2^{nd} ball can be put in any of these 3 boxes.

3^{rd} ball can be put in any of these 3 boxes.

4^{th} ball can be put in any of these 3 boxes.

i.e., these 4 balls can be arranged in

3×3×3×3=343×3×3×3=34 ways

Required number of ways

= ^{7}C_{1} × ^{6}C_{2} ×34×34

=7×15×81=8505

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In how many ways can 7 different balls be distributed in 5 different boxes if any box can contain any number of balls and no box is left empty?

A.

16800

B.

12400

C.

22000

D.

19700

Answer with explanation

Answer: Option AExplanation

Case 1: 1 box gets 3 balls, the other 4 boxes each get 1 ball

Number of box options for the box with 3 balls = 5. (Any of the 5 boxes.)

From the 7 balls, the number of ways to choose 3 balls for this box = 7C3 = (7*6*5)/(3*2*1) = 35.

Number of ways to arrange the remaining 4 balls = 4! = 24.

To combine these options, we multiply:

5*35*24 = 4200.

Case 2: 2 boxes each get 2 balls, the other 3 boxes each get 1 ball

From the 5 boxes, the number of ways to choose a pair of boxes to hold 2 balls each = 7C2 = (5*4)/(2*1) = 10.

From the 7 balls, the number of ways to choose a pair of balls for the 1st box = (7*6)/*2*1) = 21.

From the remaining 5 balls, the number of ways to choose a pair of balls for the 2nd box = (5*4)/(2*1) = 10.

Number of ways to arrange the remaining 3 balls = 3! = 6.

To combine these options, we multiply:

10*21*10*6 = 12600.

Total ways = 4200+12600 = 16800.

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A box contains 12 different black balls, 7 different red balls and 6 different blue balls. In how many ways can the balls be selected?

A.

728

B.

2^{25} – 1

C.

2^{25}

D.

727

Answer with explanation

Answer: Option BExplanation

Number of ways in which one or more objects can be selected from n distinct objects *(i.e., we can select 1 or 2 or 3 or … or n objects at a time)*

= ^{n}C_{1} + ^{n}C_{2} + … + ^{n}C_{n} = 2^{n} – 1

It is explicitly stated that 12 black balls are different, 7 red balls are different and 6 blue balls are different. Hence there are 25 (=12+ 7+ 6) different balls.

We can select one ball from 25 balls, two balls from 25 balls, … 25 balls from 25 balls.

Hence, required number of ways.

= Number of ways in which 1 ball can be selected from 25 distinct balls

+ Number of ways in which 2 balls can be selected from 25 distinct balls

+ Number of ways in which 3 balls can be selected from 25 distinct balls

…

+ Number of ways in which 25 balls can be selected from 25 distinct balls

= ^{25}C_{1} + ^{25}C_{2} + … + ^{25}C_{25}

=225−1

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How many 5 digit numbers can be formed using the digits 1,2,3,4,⋯9 such that no two consecutive digits are the same?

A.

9×84

B.

4×84

C.

95

D.

85

Answer with explanation

Answer: Option AExplanation

Here, no two consecutive digits can be the same.

The ten thousands place can be filled by any of the 9 digits (1,2,3,4,… 9)

9 |

Repetition is allowed here. Only restriction is that no two consecutive digits can be the same. Hence the digit we placed in the ten thousands place cannot be used at the thousands place. Hence thousands place can be filled by any of the 8 digits.

9 | 8 |

Similarly, hundreds place, tens place and unit place can be filled by any of the 8 digits

9 | 8 | 8 | 8 | 8 |

Hence, the required count of 5 digit numbers that can be formed using the digits 1,2,3,4,… 9 such that no two consecutive digits are same

=9×84=9×84

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How many signals can be made using 6 different coloured flags when any number of them can be hoisted at a time?

A.

1956

B.

1720

C.

2020

D.

1822

Answer with explanation

Answer: Option AExplanation

Given that any number of flags can be hoisted at a time. Hence we need to find out number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.

Number of signals that can be made using 1 flag

= ^{6}P_{1} =6=6

Number of signals that can be made using 2 flags

= ^{6}P_{2} =6×5=30=6×5=30

Number of signals that can be made using 3 flags

= ^{6}P_{3} =6×5×4=120=6×5×4=120

Number of signals that can be made using 4 flags

= ^{6}P_{4} =6×5×4×3=360=6×5×4×3=360

Number of signals that can be made using 5 flags

= ^{6}P_{5} =6×5×4×3×2=720=6×5×4×3×2=720

Number of signals that can be made using 6 flags

= ^{6}P_{6} =6×5×4×3×2×1=720=6×5×4×3×2×1=720

Therefore, required number of signals

=6+30+120+360+720+720=1956

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In how many ways can 10 software engineers and 10 civil engineers be seated around a round table so that they are positioned alternatively?

A.

9! × 10!

B.

10! × 10!

C.

2 × (10!)^{2}

D.

2 × 9! × 10!

Answer with explanation

Answer: Option AExplanation

10 civil engineers can be arranged around a round table in

(10−1)!=9!(10−1)!=9! ways …(A)

Now we need to arrange software engineers such that software engineers and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers in the 10 positions marked as * as shown below.

This can be done in 10!10! ways …(B)

From (A) and (B),

required number of ways =9!×10!

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Kiran has 8 black balls and 8 white balls. In how many ways can he arrange these balls in a row so that balls of different colours alternate?

A.

8!×7!

B.

2×8!×7!

C.

2×(8!)2

D.

(8!)2

Answer with explanation

Answer: Option CExplanation

8 black balls can be arranged in 8! ways …(A)

Now we need to arrange white balls such that white balls and black balls are positioned alternatively. i.e., we can arrange 8 white balls either in the 8 positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I as shown below.

8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H in 8! ways.

8 white balls can be arranged in the 8 positions marked as B,C,D,E,F,G,H,I in 8! ways.

8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I in 8!+8!=2×8!8!+8!=2×8! ways …(B)

From (A) and (B),

required number of ways =8!×2×8!=2×(8!)2

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In how many ways can 9 different colour balls be arranged in a row so that black, white, red and green balls are never together?

A.

146200

B.

219600

C.

314562

D.

345600

Answer with explanation

Answer: Option DExplanation

Total number of ways in which 9 different colour balls can be arranged in a row

=9! ⋯=9! ⋯(A)

Now we will find out total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are always together.

We have total 9 balls. Since black, white, red and green balls are always together, group these 4 balls together and consider as a single ball. Hence we can take total number of balls as 6. These 6 balls can be arranged in 6!6! ways.

We had grouped 4 balls together. These 4 balls can be arranged among themselves in 4!4!ways.

Hence, total number of ways in which 9 different colour balls be arranged in a row so that black, white, red and green balls are always together

=6!×4! ⋯=6!×4! ⋯(B)

From (A) and (B),

Total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are never together

=9!–6!×4!=6!×7×8×9−6!×4!=6!(7×8×9–4!)=6!(504–24)=6!×480=720×480=345600

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There are two books each of 5 volumes and two books each of two volumes. In how many ways can these books be arranged in a shelf so that the volumes of the same book should remain together?

A.

4! × 5! × 2!

B.

4! × 14!

C.

14!

D.

4! × 5! × 5! × 2! × 2!

Answer with explanation

Answer: Option DExplanation

1 book: 5 volume

1 book: 5 volume

1 book: 2 volume

1 book: 2 volume

Given that volumes of the same book should remain together. Hence, just tie the same volume books together and consider as a single book. Hence we can take total number of books as 4. These 4 books can be arranged in ^{4}P_{4} =4!=4! ways.

5 volumes of the 1^{st} book can be arranged among themselves in ^{5}P_{5} =5!=5! ways.

5 volumes of the 2^{st} book can be arranged among themselves in ^{5}P_{5} =5!=5! ways.

2 volumes of the 3^{rd} book can be arranged among themselves in ^{2}P_{2} =2!=2! ways.

2 volumes of the 4^{th} book can be arranged among themselves in ^{2}P_{2} =2!=2! ways.

Hence total number of ways

=4!×5!×5!×2!×2!

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Arun wants to send invitation letter to his 7 friends. In how many ways can he send the invitation letter if he has 4 servants to carry the invitation letters?

A.

9! × 8

B.

9!

C.

9! × 2!

D.

10! × 2!

Answer with explanation

Answer: Option AExplanation

Total number of ways in which we can arrange 10 books on a shelf

= ^{10}P_{10} =10! ⋯=10! ⋯ (A)

Now we will find out total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together.

We have a total of 10 books. If a particular pair of books must always be together, just tie these two books together and consider as a single book. Hence we can take total number of books as 9. These 9 books can be arranged in ^{9}P_{9} =9!=9! ways.

We had tied two books together. These books can be arranged among themselves in ^{2}P_{2} =2!=2! ways.

Hence, total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together

=9!×2! ⋯=9!×2! ⋯ (B)

From (A) and (B),

Total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will never be together

10!–(9!×2!)=10!–(9!×2)=(9!×10)−(9!×2)=9!(10−2)=9!×8

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Find the number of triangles that can be formed using 14 points in a plane such that 4 points are collinear?

A.

480

B.

360

C.

240

D.

120

Answer with explanation

Answer: Option BExplanation

Suppose there are n points in a plane out of which m points are collinear. Number of triangles that can be formed by joining these n points as vertices

= ^{n}C_{3} – ^{m}C_{3}

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