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Test: Permutations And Combinations - 2 - JEE MCQ


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25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Permutations And Combinations - 2

Test: Permutations And Combinations - 2 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Permutations And Combinations - 2 questions and answers have been prepared according to the JEE exam syllabus.The Test: Permutations And Combinations - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutations And Combinations - 2 below.
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Test: Permutations And Combinations - 2 - Question 1

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

Test: Permutations And Combinations - 2 - Question 2

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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 2

Sum of the numbers (S) formed by taking all the given n digits is-

S=(sum of all digits)×(n−1)!×(111....n times)

Now according to the question, the digits are 1,3,5,7

∴n = 4

Sum of digits = 1+3+5+7 = 16

∴S = 16×3!×1111

Hence the correct answer is  16×3!×1111.

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Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 3

 

Test: Permutations And Combinations - 2 - 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

Test: Permutations And Combinations - 2 - Question 5

There are 10 true-false questions. The number of ways in which they can be answered is [ it is must to attempt the question ]

Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 6

′+′ 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.

Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 7

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)

Test: Permutations And Combinations - 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?

Detailed Solution for Test: Permutations And Combinations - 2 - Question 8

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

Test: Permutations And Combinations - 2 - 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:

Detailed Solution for Test: Permutations And Combinations - 2 - Question 9

Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 10

It is asked to find out the number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated.

 

According to the question, there are 8 flowers, but it is given that four particular flowers are never separated. So four particular flowers will always remain adjacent to each other.

To solve this problem we should know the following cases of circular permutations:

  • If clockwise and anti-clockwise orders are different, the total number of circular permutations is: (n-1)!.
  • If anti-clockwise and clockwise orders are taken as the same, then the total number of circular permutations is given by (n-1)!/2!

It is important to know which formula is applicable in this case.

When we turn a garland it forms a different arrangement. But it still is the same garland.
For example, consider a set of numbers like ‘1234’. When turned around it looks like ‘4321’ but it still is the same permutation of numbers.
In a similar way, the garland when turned might have a different permutation but it still is the same garland.
Therefore we’ll use the 2nd formula.
For n distinct objects in which clockwise and anticlockwise orders are taken as the same, the total number of circular permutations is: (n-1)!/2
Here in this case, given that four flowers must be together in all cases.
So we’ll club together all four particular flowers and consider this as one object.
Now we have 5 different objects - four flowers and four particular flowers clubbed as one object. Now we have to arrange these five objects in five circular seats.
In that case, the number of permutations is (5-1)!
And in every permutation, the clubbed flowers can be arranged in 4! ways.
So there are a total of (5-1)! * 4! Permutations.
But we have seen that the garland is the same in anticlockwise and clockwise order.
Therefore for every permutation, we have included its rotated permutation.
So the total number of circular permutations of garlands is : (5-1)!*4! /2

(5-1)! * 4! /2 = (4*3*2*1)*(4*3*2*1)/2
= 24 * 12 = 288

Thus the total number of ways in which 8 different flowers can be seated to form a garland so that four particular flowers are never separated is 288.

Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 11

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.

Test: Permutations And Combinations - 2 - Question 12

The number of all odd divisors of 3600 is

Detailed Solution for Test: Permutations And Combinations - 2 - Question 12

 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.

Test: Permutations And Combinations - 2 - Question 13

The number of all even divisors of 1600 is

Test: Permutations And Combinations - 2 - Question 14

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

Test: Permutations And Combinations - 2 - Question 15

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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 15

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

Test: Permutations And Combinations - 2 - Question 16

Number of all 4 digit numbers with distinct digits is

Test: Permutations And Combinations - 2 - Question 17

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

Test: Permutations And Combinations - 2 - Question 18

An online examination have 12 question out of which we havethe alternative to select the answer. Choose how many ways can be there in which one can answer the question

Detailed Solution for Test: Permutations And Combinations - 2 - Question 18

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

Test: Permutations And Combinations - 2 - 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)

Test: Permutations And Combinations - 2 - Question 20

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

Test: Permutations And Combinations - 2 - Question 21

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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 21

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

Test: Permutations And Combinations - 2 - 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

Detailed Solution for Test: Permutations And Combinations - 2 - Question 22

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!

Test: Permutations And Combinations - 2 - Question 23

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

Test: Permutations And Combinations - 2 - Question 24

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

Test: Permutations And Combinations - 2 - Question 25

2.6.10.14……upto 50 factors is equal to

Detailed Solution for Test: Permutations And Combinations - 2 - Question 25

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!

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