Topic: Permutations and Combinations (Test 1)



Topic: Permutations and Combinations
Q.1
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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Q.2
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)

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Q.3
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.

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Q.4
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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Q.5
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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Q.6
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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Q.7
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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Q.8
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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Q.9
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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Q.10
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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