I. Single flag of different colours

II. Any two flags in a different sequence of colours.

III. three flags in a different sequence of colours.

The maximum number of codes that can be generated is.

- 1
- 2
- 3
- 4
- Next Page »

A Learning Portal from Recruitment India

Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.

A.

36

B.

48

C.

144

D.

96

Answer with explanation

Answer: Option CExplanation

Explanation: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

Workspace

Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition.

A.

666600

B.

665500

C.

666700

D.

678860

Answer with explanation

Answer: Option AExplanation

Explanation: The given digits are 1, 3, 5, 7, 9

Sum of r digit number= n-1Pr-1

(Sum of all n digits)×(1111… r times)

N is the number of non zero digits.

Here n=5, r=4

The sum of 4 digit numbers

4P3 (1+3+5+7+9)(1111)=666600.

Workspace

In a colony, there are 55 members. Every member posts a greeting card to all the members. How many greeting cards were posted by them?

A.

2970

B.

990

C.

1980

D.

890

Answer with explanation

Answer: Option AExplanation

Explanation: First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are

54 + 54 + 54 …

54 (55times) = 54 x 55 = 2970.

Workspace

How many three letter words are formed using the letters of the word TIME?

A.

12

B.

16

C.

20

D.

24

Answer with explanation

Answer: Option DExplanation

Explanation:

The number of letters in the given word is four.

The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

Workspace

In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?

A.

5040

B.

480

C.

720

D.

360

Answer with explanation

Answer: Option CExplanation

The word ‘LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Workspace

In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

A.

288

B.

12

C.

256

D.

144

Answer with explanation

Answer: Option DExplanation

Solution:

Let the Arrangement be,

**B G B G B G B**

4 boys can be seated in 4! Ways

Girl can be seated in 3! Ways

Required number of ways,

= 4! × 3!

= 144

Workspace

In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

A.

9 × 8!

B.

7 × 9!

C.

8 × 8!

D.

8 × 9!

Answer with explanation

Answer: Option DExplanation

Solution:

No. of ways in which 10 paper can arranged is 10! Ways.

When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.

These 9 papers can be arranged in 9! Ways.

And two papers can be arranged themselves in 2! Ways.

No. of arrangement when best and worst paper do not come together,

= 10! – 9! × 2!

= 9!(10 – 2)

= 8 × 9!

Workspace

How many different words can be formed using all the letters of the word ALLAHABAD?

(a) When vowels occupy the even positions.

(b) Both L do not occur together.

A.

7560,60,1680

B.

7650,200,4444

C.

None of these

D.

7890,120,650

Answer with explanation

Answer: Option CExplanation

Workspace

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?

A.

3! 4! 8! 4!

B.

4! 4!

C.

3! 8!

D.

8! 4! 4!

Answer with explanation

Answer: Option AExplanation

Solution:

Taking all person of same nationality as one person, then we will have only three people.

These three person can be arranged themselves in 3! Ways.

8 Indians can be arranged themselves in 8! Way.

4 American can be arranged themselves in 4! Ways.

4 Englishman can be arranged themselves in 4! Ways.

Hence, required number of ways = 3! 8! 4! 4! Ways.

Workspace

A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?

A.

119

B.

29295

C.

209

D.

(4!-1) × (5!-1) × (6!-1)

Answer with explanation

Answer: Option BExplanation

At least 1 question from each section is compulsory, so from the 1^{st} section the candidate can attempt 1 or 2 or 3 or 4 questions.

In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.

So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.

As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2^{4} – 1

Similarly for the 2^{nd} section there are 2^{5} – 1 ways in which he can attempt and for the 3^{rd} section there are 2^{6} – 1 ways.

The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.

Thus, total number of ways in which he can attempt questions in that paper:

= (2^{4} – 1)(2^{5} – 1)(2^{6} – 1)

= 15 × 31 × 63

= 29295

Workspace

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

A.

B.

30

C.

32

D.

31

Answer with explanation

Answer: Option CExplanation

The 5 letter word can be rearranged in 5! = 120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then, the 25th word will start will CA _ _ _ .

The remaining 3 letters can be rearranged in 3! = 6 Ways.

i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.

The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24 + 6 + 2 = 32

Workspace

There are two identical red, two identical black, and two identical white balls. In how many different ways can the balls be placed in the cells (Each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

A.

15

B.

18

C.

24

D.

30

Answer with explanation

Answer: Option CExplanation

Case I : 2 balls of the same colour and two balls are a different colour are arranged.

Two balls of the same colour and two balls of different colours can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways

Therefore, Total number of arrangements = 6×3 =18 ways

Case II : Two colours out of 3 can be selected in = 3C1 = 3ways

Now 2 balls of each colour can be arranged alternatively in 2 ways

Thus 4 balls can be arranged(two of each colours)

= 3×2 = 6ways

Hence total number of arrangements = 18+6 =24 ways

Workspace

There are 8 men and 7 women. In how many ways a group of 5 people can be made such that at least 3 men are there in the group?

A.

1844

B.

1626

C.

1722

D.

1545

Answer with explanation

Answer: Option CExplanation

Case 1: 3 men and 2 women

^{8}C_{3}*^{7}C_{2} = 1176

Case 2: 4 men and 1 women

^{8}C_{4}*^{7}C_{1} = 490

Case 3: all 5 men

^{8}C_{5} = 56

Add all the cases.

Workspace

A.

70

B.

35

C.

105

D.

210

Answer with explanation

Answer: Option AExplanation

All the boxes contain distinct number of chocolates.

For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is,

^{8}C_{4} = 70

Workspace

How many distinguishable permutations of the letters in the word BANANA are there?

A.

720

B.

120

C.

60

D.

360

Answer with explanation

Answer: Option CExplanation

In BANANA we have six letters in total but here we have some duplicate letters too so we have to deal with it and have to remove those duplicate case. B – 1 A – 3 N – 2 So total no of words possible is factorial(6) ie 6! but we must remove duplicate words: ie- (6!/(2!*3!)) which gives 60 So 60 distinguishable permutation of the letters in BANANA

Workspace

George started a hardware business by investing Rs. 50,000. After six months, Kate joined him with a capital of Rs. 70,000. After 3 years, they earned a profit of Rs. 25,000. Find George’s share in profit.

A.

Rs. 13000.00

B.

Rs. 13224.65

C.

Rs. 12500.50

D.

Rs. 11538.46

Answer with explanation

Answer: Option DExplanation

Workspace

In how many ways can 20 boys and 18 girls make a queue such that no two girls are together?

A.

20!* ^{20}C_{18}

B.

20!* ^{20}P_{18}

C.

20!* ^{21}C_{18}

D.

20!* ^{21}P_{18}

Answer with explanation

Answer: Option BExplanation

The boys will be arranged in 20! ways. Now, there are a total of 21 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls). Therefore, the 18 girls can stand at these 21 places only. Hence, the number of ways = 20!* ^{21}P_{18}

Workspace

I. Single flag of different colours

II. Any two flags in a different sequence of colours.

III. three flags in a different sequence of colours.

The maximum number of codes that can be generated is.

A.

6

B.

9

C.

15

D.

18

Answer with explanation

Answer: Option CExplanation

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

We can choose any colour.

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

Workspace

A box contains three white balls, four black balls and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is?

A.

50

B.

100

C.

150

D.

200

Answer with explanation

Answer: Option BExplanation

The required number of ways

(a) 1 black and 2 others = ^{4}C_{1}.^{6}C_{2} = 4 × 15 = 60

(b) 2 black and 1 other = ^{4}C_{2}.^{6}C_{1} = 6 × 6 = 36

(c) All the three black = ^{4}C_{3} = 4

Total =60 + 36 + 4 = 100

Workspace

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

A.

17756

B.

17576

C.

12657

D.

12666

Answer with explanation

Answer: Option BExplanation

Workspace

A.

5

B.

6

C.

7

D.

8

Answer with explanation

Answer: Option CExplanation

Number of ways of opting a subject other than Mathematics II = 4C2. = 4x3x2!/2!x2 = 6.

Number of ways of selection of Mathematics II = 1

Therefore, Total Number of ways = 6+1 =7.

Workspace

A.

252

B.

250

C.

1001

D.

Answer with explanation

Answer: Option CExplanation

The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5 – 1) four walls in the single row, which can be done in ^{14}C_{4} ways. (The balls placed between every successive pair of walls belong to one group) ^{14}C_{4} = 1001 ways.

Workspace

In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is

A.

7

B.

9

C.

10

D.

None of these

Answer with explanation

Answer: Option AExplanation

Let n be the number of teams.

^{n}C_{2} = 21

(n(n-1)/2) = 21

⇒ n(n-1) = 42 ∴

⇒ n = 7

Workspace

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

A.

601

B.

600

C.

603

D.

602

Answer with explanation

Answer: Option AExplanation

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Workspace

- 1
- 2
- 3
- 4
- Next Page »

Correct Answer 👍

Wrong Answer 👎