In a multiple choice question, there are 4 alternatives, of which one or more are correct. The number of ways in which a candidate can attempt this question is

**A. ** 4

**B. ** 15

**C. ** 25

**D. ** 16

**Answer : ****Option B**

**Explaination / Solution: **

Since given there are four alternatives in which one or more are correct,we have to consider the following four cases

The candidate choose 1 correct answer, 2 correct answers,3 correct answers or 4 correct answers.

1 correct answer can be chosen in 4C1 ways = 4 ways

2 correct answer can be chosen in 4C2 ways= 6 ways

3 correct answers can be chosen in 4C3ways = 4 ways

4 correct answers can be chosen in4C4 ways = 1 way

Hence the totalnumber of ways = 4 + 6 + 4+ 1=15ways

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The number of arrangements of n different things taken r at a time which include a particular thing is

**A. ** n P(n-1,r)

**B. ** P(n-1, r-1 )

**C. ** None of these

**D. ** r P (n-1,r-1)

**Answer : ****Option D**

**Explaination / Solution: **

The number of arrangements of n different things taken r at a time which include a particular thing is r n-1Pr-1 = rP(n-1,r-1)

The number of arrangements of n different things taken r at a time which include a particular thing is r n-1Pr-1 = rP(n-1,r-1)

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A fair dice is rolled n times. The number of all the possible outcomes is

**A. ** None of these

**B. ** 6n

**C. ** 6n
**D. ** n6
**Answer : ****Option C**

**Explaination / Solution: **

each time there are 6 possibilities, therefore for n times there are 6n possibilities.

each time there are 6 possibilities, therefore for n times there are 6n possibilities.

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The number of different ways in which a man can invite one or more of his 6 friends to dinner is?

**A. ** 15

**B. ** 63

**C. ** 120

**D. ** 30

**Answer : ****Option B**

**Explaination / Solution: **

He can invite any one friend in 6C1 ways= 6 ways:

He can invite any two friends in 6C2 ways = 15 ways

He can invite any three friends in 6C3 ways =20 ways

He can invite any 4 friends in 6C4ways = 15 ways

He can invite any 5 friends in 6C5 ways = 6 ways

He can invite all the 6 friends in 6C6 ways= 1 way.

Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.

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The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is

**A. ** 370

**B. ** 365

**C. ** None of these

**D. ** 360

**Answer : ****Option B**

**Explaination / Solution: **

You have two different kinds of such three-digit even numbers.First is 5 at the hundred'splace and second 5 is not at the hundred'splace

In first case no is of the form 57x, where x is the unit's digit ,which can be 0,2,4,6,8 which is just 5 possibilities.Hence the no of possibilities in this case is 1x1x5=5

In second case the hundred's digit can be 1,2,3,,4,,6,7,8,or,9 which is 8 ways and the ten's digit can be any of the 9 numbers and unit digit can be any of the 5 even numbers .Therfore the no: of ways will be 8×9×5=360

So total we have 360 + 5 = 365 possibilities.

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The number of distinguishable ways in which the 4 faces of a regular tetrahedron can be painted with 4 different colours is

**A. ** 4

**B. ** 24

**C. ** 2

**D. ** None of these

**Answer : ****Option C**

**Explaination / Solution: **

We have a regular tetrahedron has 4 faces and we have to colour it with 4 different colours in ways.

But in this we will be getting many overcountings .

We have there are 12 ways in which we can orient a regular tetrahedron

Hence the number of distinct ways of colouring a regular tetrahedron with 4 different colours is

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The number of ways in which 6 “ + “ and 4 “ – “ signs can be arranged in a line such that no two “ – “ signs occur together is

**A. ** 5040

**B. ** 210

**C. ** 35

**D. ** 120

**Answer : ****Option C**

**Explaination / Solution: **

Since all the plus signs are identical , we have number of ways in which 6 plus signs can be arranged=1.

Now we will have 7 empty slots between these 6 identical + signs

Hence number of possible places of - sign =7

Therefore number of ways in which the 4 minus sign can take any of the possible 7 places=

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Different calendars for the month of February are made so as to serve for all the coming years. The number of such calendars is

**A. ** 7

**B. ** 2

**C. ** 14

**D. ** 8

**Answer : ****Option C**

**Explaination / Solution: **

The month of February has either 28 days (non-Leap Year) or 29 days(Leap Year).

Thus there are 2 possibility for the number of days in February.

Also the first day of the month of February can be any of the 7 days in a week.

So total number of February calendars = 2 * 7 =14

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The number of all possible positive integral solutions of the equation xyz = 30 is

**A. ** 27

**B. ** 26

**C. ** None of these

**D. ** 25

**Answer : ****Option A**

**Explaination / Solution: **

Given xyz=30

We have the possible values of x ,y,z are the following triads

1,1,30

1,2,15

1,3,10

1,5,6

2,3,5

First one can have 3!/2! = 3 ways and the remaining four triads can have 3! combinations

Hence total combinations = 3 + 4*3! = 27

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The number of ways of dividing 52 cards equally into 4 sets is

**A. ** 52!(13!)4
**B. ** 52!4!
**C. ** 52!4!(13!)4
**D. ** 52!4(13!)4
**Answer : ****Option C**

**Explaination / Solution: **

If 4n things are to form 4 equal groups then the numbers of ways are

Hence number of ways to divide 52 cards to form 4 groups of 13 cards each =

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