Test: Permutations And Combinations (CBSE Level) - 2


25 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Permutations And Combinations (CBSE Level) - 2


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This mock test of Test: Permutations And Combinations (CBSE Level) - 2 for JEE helps you for every JEE entrance exam. This contains 25 Multiple Choice Questions for JEE Test: Permutations And Combinations (CBSE Level) - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Permutations And Combinations (CBSE Level) - 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Permutations And Combinations (CBSE Level) - 2 exercise for a better result in the exam. You can find other Test: Permutations And Combinations (CBSE Level) - 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

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

Solution:
QUESTION: 2

The sum of all the numbers which can be formed by using the digits 1 , 3 , 5 , 7 ,9 all at a time and which have no digit repeated is

Solution:
QUESTION: 3

4 boys and 4 girls are to be seated in a row. The number of ways in which this can be done, if the boys and girls sit alternately, is

Solution:
QUESTION: 4

The letters of the word ‘SOCIETY’ are arranged in such a manner that the vowels and consonants occur alternately, the number of different words so obtained is

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QUESTION: 5

There are 10 true-false questions. The number of ways in which they can be answered is

Solution:
QUESTION: 6

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

Solution:

′+′ signs can be put in a row in one way creating seven gaps shown as arrows:
Now 4′−′ signs must be kept in these gaps. So, no tow ′−′ signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways.

QUESTION: 7

On a railway track, there are 20 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is

Solution:

Number of tickets selected from first station =20
from second =19
.... for last station =0
We have to select 2 consecutive stations
so total number of possible tickets = P(20,2)

QUESTION: 8

A class is composed 2 brothers and 6 other boys. In how many ways can all the boys be seated at the round table so that the 2 brothers are not seated besides each other?

Solution:

Take 1 person from 6 and fix him and 5 others can arranged in -- 5! ways=120

there are 6 places left in which 2 brothers can sit

so they can choose any 2 places from 6 - 6C2 ways=15

2 brothers can arrange themselves in 2! ways=15*2=30

total ways=120*30=3600

QUESTION: 9

The number of all selections which a student can make for answering one or more questions out of 8 given questions in a paper, when each question has an alternative, is:

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QUESTION: 10

The number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated is

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QUESTION: 11

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

Solution:

The mint has to perform two jobs:
1) Selecting the number of days in the February month i.e., 28 or 29
2) Selecting the first day of February month.
The first job can be completed in 2 ways while second job can be completed in 7 ways by selecting any one of the seven days of a week.
Thus, the required number of plates = 2 x 7 = 14.

QUESTION: 12

The number of all odd divisors of 3600 is

Solution:

 Number 1 is odd. As any number is divisible by 1: 1
Do prime factorisation of 3600: 2*2*2*2*3*3*5*5
Select all odd numbers from above: 3,3,5,5
Try every possible products of these:
Single number: 3, 5
Two numbers: 3*3, 3*5, 5*5: 9,15,25
Three numbers: 3*3*5, 3*5*5: 45, 75
All four: 3*3*5*5: 225
The odd divisors are: 1,3,5,9,15,25,45,75,225
There are 9 odd divisors of 3600.

QUESTION: 13

The number of all even divisors of 1600 is

Solution:
QUESTION: 14

A convex polygon of n sides has n diagonals. The value of n is

Solution:
QUESTION: 15

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

Solution:

All possible three number multiplications originate from 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
total combinations = 3 + 4*3! = 27

QUESTION: 16

Number of all 4 digit numbers with distinct digits is

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QUESTION: 17

The number of ways, in which a student can choose 5 courses out of 8 courses, when 2 courses are compulsory, is

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QUESTION: 18

The number of ways, in which a student can select one or more questions out of 12 each having an alternative, is

Solution:

No.of choices for each question = 3
 
There are six questions so total = (3)12
So no.of ways = (3)12 - 1

QUESTION: 19

20 students can compete for a race. The number of ways in which they can win the first three places is (given that no two students finish in the same place)

Solution:
QUESTION: 20

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

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QUESTION: 21

The number of three digit numbers having atleast one digit as 5 is

Solution:

These digit number without digit 5 →100....999
→ these are 900 three-digit number
→ from 100 to 199 → 19 number with 5.
200−299→19
300−399→19
400−499→19
600−699→19
700−799→19
800−899→19
900−999→19
500−599→100
total number with 5=19×8+100 for (500-599)
 =152+100
 =252

QUESTION: 22

5 boys and 5 girls are to be seated around a table such that boys and girls sit alternately. The number of ways of seating them is

Solution:

First we fix the alternate position of the girls. Five girls can be seated around the circle in (5−1)!=4! , 5 boys can be seated in five -vacant place by 5!
∴ Required number of ways =4!×5!

QUESTION: 23

The total number of numbers from 1000 to 9999 (both inclusive) that do not have 4 different digits

Solution:
QUESTION: 24

If P (n,r) = C (n,r) then

Solution:
QUESTION: 25

2.6.10.14……upto 50 factors is equal to

Solution:

2.6.10.14... to n factors = (2n)!/n! 
= 2(1.3.5.7………………25)
2.6.10.14... to 50 factors = (2*2*25)!/25! 
 = 100!/25!