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In how many ways can 5 men and 2 ladies be arranged at a round table if two ladies are never together?

A.

720

B.

420

C.

580

D.

480

Answer with explanation

Answer: Option DExplanation

total no. of arrangements = 6! = 720

No. of arrangements in which 2 ladies are together = 2 X 5! = 240

And the arrangements in which two ladies are never together = 720 – 240 = 480

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A box contains 4 red, 3 white, and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls in different colors

A.

168

B.

12

C.

24

D.

48

Answer with explanation

Answer: Option CExplanation

This question seems to be a bit typical, isn’t, but it is simplest.

1 red ball can be selected in 4C1 ways

1 white ball can be selected in 3C1 ways

1 blue ball can be selected in 2C1 ways

Total number of ways

= 4C1 x 3C1 x 2C1

= 4 x 3 x 2

= 24

Please note that we have multiplied the combination results, we use to add when there is OR condition, and we use to multiply when there is AND condition, In this question it is AND as

1 red AND 1 White AND 1 Blue, so we multiplied.

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How many 3 digit numbers are divisible by 4?

A.

125

B.

158

C.

225

D.

252

Answer with explanation

Answer: Option CExplanation

A number is divisible by 4 when its last two digits are divisible by 4

For this, the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96

By the formula, a_{n} = a + (n-1)d

====>>>> 96 = 0 + (n-1)*4

==== >>>> n = 25

so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit

====>>>> so total 9*25 = 225

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Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

A.

310

B.

320

C.

330

D.

340

Answer with explanation

Answer: Option CExplanation

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

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Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?

A.

8

B.

6

C.

5

D.

7

Answer with explanation

Answer: Option DExplanation

Maximum number of such different groups = ABC, ABD, ABE, BCE,BDE,CEA,DEA =7.

Alternate method:

Total number of way in which 3 boys can be selected out of 5 is 5C_{3}

Number of ways in which CD comes together = 3 (CDA,CDB,CDE)

Therefore, Required number of ways = 5C_{3} -3

===>>>>>>=== 10-3 = 7.

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There are 5 floating stones on a river. A man wants to cross the river. He can move either 1 or 2 steps at a time. Find the number of ways in which he can cross the river?

A.

13

B.

12

C.

14

D.

11

Answer with explanation

Answer: Option AExplanation

The man needs to take 6 steps to cross the river. He can do this in the following ways:

===>>> Crossing the river by 6 unit steps = 1 way.

===>>> Crossing the river by 4 unit steps and 1 double step = ^{5}C_{1} = 5C4 = 5 ways.

===>>> Crossing the river by 2 unit steps and 2 double steps = ^{4}C_{2} = 6 ways.

===>>> Crossing the river by 3 double steps = 1 way.

Hence, the required number of ways = 1 + 5 + 6 + 1 = 13.

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How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?

A.

1000

B.

100

C.

50

D.

25

Answer with explanation

Answer: Option BExplanation

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

=> Therefore, using ‘n’ alphabets we can form n * n * n = n^{3} distinct 3 digit initials.

=> Note distinct initials is different from initials where the digits are different.

=> For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

=>This n^{3} different initials = 1 million

======>>>>> i.e. n^{3} = 10^{6} (1 million = 10^{6})

======>>>>> n^{3} = (10^{2})^{3} => n = 10^{2} = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

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Nine chairs are numbered 1 to 9. Three women and four men wish to occupy one chair each. First, the women chose the chairs from amongst the chair marked 1 to 5; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

A.

^{5}C_{2} × ^{4}P_{3}

B.

^{5}C_{3} × ^{6}C_{4}

C.

^{5}C_{3} × ^{4}C_{2}

D.

None of these

Answer with explanation

Answer: Option BExplanation

Women can select 3 chairs from chairs numbered 1 to 5 in ^{5}C_{3} ways and remaining 6 chairs can be selected by 4 men in ^{6}C_{4} ways. Hence the required number of ways = ^{5}C_{3} × ^{6}C_{4}

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How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

A.

360

B.

720

C.

1420

D.

1680

Answer with explanation

Answer: Option DExplanation

The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.

Given that no digit appears more than once. Hence we have 8 digits remaining (0,1,2,4,6,7,8,9)(0,1,2,4,6,7,8,9)

So, the next 4 places can be filled with the remaining 8 digits in ^{8}P_{4} ways.

Total number of ways = ^{8}P_{4} =8×7×6×5=1680

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How many different four letter words can be formed (the words need not be meaningful using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?

A.

53

B.

55

C.

59

D.

69

Answer with explanation

Answer: Option CExplanation

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

**Case 1**: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

**Case 2**: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

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