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.

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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

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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

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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

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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.

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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

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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

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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

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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}

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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

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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

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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

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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.

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