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In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?

A.

120

B.

240

C.

360

D.

480

Answer with explanation

Answer: Option DExplanation

As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

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A group consists of 4 men, 6 women and 5 children. In how many ways can 3 men, 2 women and 3 children selected from the given group?

A.

300

B.

450

C.

600

D.

750

Answer with explanation

Answer: Option CExplanation

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃

= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)

= 4 * 15 * 10

= 600 ways.

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The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is

A.

4000

B.

2160

C.

4320

D.

5300

Answer with explanation

Answer: Option CExplanation

The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.

The required number of ways = 6(6!) = 4320

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In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?

A.

25

B.

13

C.

40

D.

30

Answer with explanation

Answer: Option DExplanation

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.

3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

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How many four digit numbers can be formed using the digits {1, 3, 4, 5, 7,9}(repetition of digits is not allowed)?

A.

360

B.

60

C.

300

D.

180

Answer with explanation

Answer: Option AExplanation

The given digits are six.

The number of four digit numbers that can be formed using six digits is ⁶P₄ = 6 * 5 * 4 * 3 = 360.

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A committee has 5 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee?

A.

150

B.

200

C.

250

D.

300

Answer with explanation

Answer: Option BExplanation

The number of ways to select two men and three women = ⁵C₂ * ⁶C₃

= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)

= 200

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Using all the letters of the word “THURSDAY”, how many different words can be formed?

A.

8

B.

8!

C.

7!

D.

7

Answer with explanation

Answer: Option BExplanation

Total number of letters = 8

Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.

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A.

24

B.

23

C.

21

D.

18

Answer with explanation

Answer: Option CExplanation

__Case 1: Exactly one box contains a blue ball__

One blue ball can be placed into any of the 6 boxes. i.e, 6 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways =6×1=6=6×1=6

__Case 2: Exactly two boxes contain blue balls__

Two blue balls can be placed into (box 1 and box 2) or (box 2 and box 3) or (box 3 and box 4) or (box 4 and box 5) or (box 5 and box 6). i.e, 5 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways =5×1=5=5×1=5

__Case 3: Exactly three boxes contain blue balls__

Three blue balls can be placed into (box 1 , box 2 and box 3) or (box 2, box 3 and box 4) or (box 3 , box 4 and box 5) or (box 4, box 5 and box 6). i.e, 4 ways of doing this.

Red balls can be filled in the remaining boxes. Since red balls are identical, there is only 1 way of doing this.

Total number of ways =4×1=4=4×1=4

__Case 4: Exactly four boxes contain blue balls__

Four blue balls can be placed into (box 1 , box 2, box 3 and box 4) or (box 2, box 3, box 4 and box 5) or (box 3 , box 4, box 5 and box 6). i.e, 3 ways of doing this.

Total number of ways =3×1=3=3×1=3

__Case 5: Exactly five boxes contain blue balls__

Five blue balls can be placed into (box 1, box 2, box 3, box 4 and box 5) or (box 2, box 3, box 4, box 5 and box 6). i.e, 2 ways of doing this.

Total number of ways =2×1=2=2×1=2

__Case 6: All the six boxes contain blue balls__

Six blue balls can be placed into (box 1 , box 2, box 3, box 4, box 5 and box 6). i.e, only 1 way of doing this.

Total number of ways = 1

Hence, required number of ways

=6+5+4+3+2+1=21

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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

Workspace

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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