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There are 4 oranges, 5 apples and 6 mangoes in a basket. In how many ways can a person make a selection of fruits among the fruits in the basket?

A.

210

B.

209

C.

256

D.

220

Answer with explanation

Answer: Option BExplanation

Number of ways in which one or more objects can be selected out of S_{1} alike objects of one kind, S_{2} alike objects of second kind , S_{3} alike objects of third kind and so on … S_{n}alike objects of n^{th} kind

= (S_{1} + 1) (S_{2} + 1)(S_{3} + 1)…(S_{n} + 1) – 1

Whenever it is not explicitly mentioned that fruits are distinct, we take them as identical. (Persons/men/women are normally considered as distinct.)

Hence, here we consider that

all 4 oranges are identical,

all 5 apples are identical

all 6 mangoes are identical.

As per the formula, required number of ways=(4+1)(5+1)(6+1)–1=5×6×7–1=209

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A.

40

B.

200

C.

180

D.

120

Answer with explanation

Answer: Option BExplanation

Let total number of women =w=w

total number of men =m=m

Number of games in which both players were women =45

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In how many ways can 10 engineers and 4 doctors be seated at a round table if no two doctors sit together?

A.

10!×4!

B.

9!×9!× ^{10}P_{4}

C.

10!×10!× ^{10}P_{4}

D.

13!

Answer with explanation

Answer: Option BExplanation

No two doctors sit together. Hence, let’s initially arrange the 10 engineers at a round table. Number of ways in which this can be done =(10−1)!=9!=(10−1)!=9! …(A)

Now there are 10 positions left (marked as *) to place the 4 doctors as shown below so that no two doctors can sit together.

Number of ways in which this can be done

= ^{10}P_{4} …(B)

From (A) and (B),

Required number of ways =9!×=9!× ^{10}P_{4}

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In how many ways can seven ‘#’ symbol and five ‘*’ symbol be arranged in a line so that no two ‘*’ symbols occur together?

A.

44

B.

56

C.

62

D.

28

Answer with explanation

Answer: Option BExplanation

There are 7 identical ‘#’ symbols and 5 identical ‘*’ symbols.

We need to arrange these 12 symbols in a line so that no two ‘*’ symbols occur together.

The seven ‘#’ symbols can be arranged in 1 way …(A)

(because all these symbols are identical and order is not important).

Now there are 8 positions to arrange the five ‘*’ symbols so that no two ‘*’ symbols occur together as indicated in the diagram below.

The five ‘*’ symbols can be placed in these 8 positions in ^{8}C_{5} ways …(B)

(because all these symbols are identical and order is not important)

From(A) and (B),

required number of ways

= 1 × ^{8}C_{5} = ^{8}C_{5}

= ^{8}C_{3} [∵ ^{n}C_{r} = ^{n}C_{(n – r)}]

=8×7×63×2×1=8×7=56

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Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes so that no box remains empty, if all balls and boxes are identical?

A.

1

B.

4

C.

2

D.

6

Answer with explanation

Answer: Option CExplanation

Here n = 3, k =5.

Hence, as per the above formula, required number of ways

= P(k,n) = P(5,3)(5,3)

The partitions of 5 into 3 parts are

1+1+31+2+21+1+31+2+2

Therefore, number of partitions of 5 into 3 parts =2=2

=> P(5,3)(5,3) =2=2

=> Required number of ways =2

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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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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,⋯91,2,3,4,⋯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 BExplanation

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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