Test: Permutations And Combinations - 1 - JEE MCQ

# Test: Permutations And Combinations - 1 - JEE MCQ

Test Description

## 25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Permutations And Combinations - 1

Test: Permutations And Combinations - 1 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Permutations And Combinations - 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Permutations And Combinations - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutations And Combinations - 1 below.
Solutions of Test: Permutations And Combinations - 1 questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Permutations And Combinations - 1 solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Permutations And Combinations - 1 | 25 questions in 25 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Permutations And Combinations - 1 - Question 1

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

Detailed Solution for Test: Permutations And Combinations - 1 - Question 1

He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

Test: Permutations And Combinations - 1 - Question 2

### A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 2

Let us say that the two particular friends are A and B.
If A is invited among six guests and B is not, then:  number of  combinations to select 5 more guests from the remaining 8 friends:
C(8, 5) =  8 ! / (5! 3!)  = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests =  C(8, 5) =  56
Suppose both A and B are not included in the six guests list : then the number of such combinations =  C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected =  140.

 1 Crore+ students have signed up on EduRev. Have you?
Test: Permutations And Combinations - 1 - Question 3

### In how many ways can a mixed doubles tennis game be arranged from a group of 10 players consisting of 6 men and 4 women

Detailed Solution for Test: Permutations And Combinations - 1 - Question 3

Mixed doubles includes 2 men and 2 women.
Since 2 men are selected of 6 men
⇒ number of ways = 6C2
Also 2 women are selected of 4 women
⇒ number of ways = 4C2
But they also can be interchanged in 2 ways
∴ Total no. of ways

Test: Permutations And Combinations - 1 - Question 4

The number of significant numbers which can be formed by using any number of the digits 0, 1, 2, 3, 4 but using each not more than once in each number is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 4

One digit positive integers are 0,1,2,3,4 =5 positive integers
Number of one digit no. can be formed = 4
Number of two digit no. can be formed = 4*4 = 16
Number of three digit no. can be formed = 4*4*3 = 48
Number of  four digit no. can be formed = 4*4*3*2 = 96
Number of five digit no. can be formed = 4*4*3*2*1 = 96
Total = 4+ 16+ 48+ 96+ 96 = 260

Test: Permutations And Combinations - 1 - Question 5

Numbers greater than 1000 but not greater than 4000 are to be formed with the digits 0, 1, 2, 3, 4, allowing repetitions, the number of possible numbers is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 5

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left-most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

So, The left-most digits have values 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

So, the next 3 digits have values 5.

Hence, there are numbers.

or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

Test: Permutations And Combinations - 1 - Question 6

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

Detailed Solution for Test: Permutations And Combinations - 1 - Question 6

Correct Answer :- c

Explanation : As given condition is , 5 must be followed by 7. So only possible way is 57X where X denotes 0,2,4,6 and 8.

So,total no. of ways=1×1×5=5

Hence, total ways in which we can make a 3-digit even no. without violating given condition are:

360+5=365

Test: Permutations And Combinations - 1 - Question 7

A coin is tossed n times, the number of all the possible outcomes is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 7

If coin is tossed n times then possible number of outcomes = 2n​

Test: Permutations And Combinations - 1 - Question 8

The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 8

There are in total 6 numbers, 4,5,6,7,8.
Now consider the number 56000
Consider the numbers of the form
56−−−.
Considering no repetitions we get
3×2×1 = 6 numbers.
Similarly for 57−−− and 58−−−.
Hence 3×6 = 18 numbers.
Now consider the numbers starting with 6.
6−−−−
We get 4×3×2×1 = 24.
Similarly for the numbers starting with 7 and 8 we get in total 24 numbers each.
Hence total number of numbers greater than 56000 will be
= (24×3)+18
= 72+18 = 90

Test: Permutations And Combinations - 1 - Question 9

If nP5 = 60n−1P3, then n is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 9

N!/(n-5)! = 60×(n-1)!/(n-1-3)!
n!/(n-5)! = 60×(n-1)!/(n-4)!
n(n-1)!/(n-5)!=60×(n-1)!/(n-4)×(n-5)!
n=60/(n-4)
n(n-4)=60
n^2-4n-60=0
(n-10)(n+6)=0
n=10 and n is not equal to -6.

Test: Permutations And Combinations - 1 - Question 10

The number of ways in which the 6 faces of a cube can be painted with 6 different colours is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 10

Number of sides of a cube = 6
Number of colour of a cube = 6
Number of ways of colouring =?
Consider a cube in 3D space
Each of the sides towards East West North South and remaning two up and down
Now fixing up positions down is also fixed
Total number of ways
= 6!/(4×6)
​= 5×3×2
= 30 Ways
Hence, the correct answer is 30 Ways

Test: Permutations And Combinations - 1 - Question 11

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

Detailed Solution for Test: Permutations And Combinations - 1 - Question 11

The correct answer is A

Given are four faces and four different colours

So , number of colours for first face=4

No. of colours for 2nd face=3 (as one will be used by the first face)

No. of colours for 3rd face=2

No. of colours for 4th face=1

So total options are=4*3*2*1=4!=24

Test: Permutations And Combinations - 1 - Question 12

The number of ways in which n ties can be selected from a rack displaying 3n different ties is

Test: Permutations And Combinations - 1 - Question 13

Find Rank of word ‘wife ‘among the words that can be formed with its letters and arranged as in dictionary is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 13

No of word formed taking E at first place = 3!
Taking F at first place = 3!
Taking I in the first place = 3!
W in first and E in second = 2!
W in first F in second and E in third = 1!
W in first I in second and E in third = 1
W in first I in second and F in third = 1
W in first I in second and F in third and E in fourth = 1
Total =24

Test: Permutations And Combinations - 1 - Question 14

The number of even numbers that can be formed by using all the digits 1, 2, 3, 4, and 5 (without repetitions) is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 14

Test: Permutations And Combinations - 1 - Question 15

The number of all numbers that can be formed by using some or all of the digits 1, 3, 5, 7, 9 (without repetitions) is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 15

Out of 1, 3, 5, 7, 9
No. of 1-digit numbers = 5
No. of 2-digit numbers = 5*4 = 20
No. of 3-digit numbers = 5*4*3 = 60
No. of 4-digit numbers = 5*4*3*2 = 120
No. of 5-digit numbers = 5*4*3*2*1 = 120
Total no. of numbers = 5 + 20 + 60 + 120 + 120 = 325

Test: Permutations And Combinations - 1 - Question 16

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

Detailed Solution for Test: Permutations And Combinations - 1 - Question 16

The total number of ways of ticking the answers in any one attempt 24 − 1 = 15.

Test: Permutations And Combinations - 1 - Question 17

In how many ways can the letters of the word ‘MATHEMATICS ‘be permuted so that consonants always occur together?

Test: Permutations And Combinations - 1 - Question 18

The number of triangles that can be formed with 6 points on a circle is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 18

Solution :- Number of such triangle = 6C3

= 6!/3!3!

=20

Test: Permutations And Combinations - 1 - Question 19

Number of ways in which 10 different things can be divided into two groups containing 6 and 4 things respectively is

Detailed Solution for Test: Permutations And Combinations - 1 - Question 19

Forming the first group by choosing 4 things out of 10 things,the total number of ways will be =10C4
Now,forming these group by choosing 6 things,the total number of ways =
6C6​
Therefore,the total number of ways = 10C4​∗6C6
= C(10,4)

Test: Permutations And Combinations - 1 - Question 20

The number of selections of n different things taken r at a time which exclude a particular thing is

Test: Permutations And Combinations - 1 - Question 21

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

Detailed Solution for Test: Permutations And Combinations - 1 - Question 21

Total Permutations of n different things taken r at a time when a particular item is always included in the arrangement is : r * (n-1 p r-1).

Test: Permutations And Combinations - 1 - Question 22

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

Test: Permutations And Combinations - 1 - Question 23

5 persons board a lift on the ground floor of an 8 storey building. In how many ways can they leave the lift?

Detailed Solution for Test: Permutations And Combinations - 1 - Question 23

Beside the ground floor, there are seven floors.
The total number of ways in which each of the five persons can leave cabin at any of the 7 floors =75
And the favorable number of ways, that is, the number of ways, in which 5 persons leave at different floors is 7P5

Test: Permutations And Combinations - 1 - Question 24

If C(12,4) + C(12,5) = C(n,5) ,then n is equal to

Test: Permutations And Combinations - 1 - Question 25

The number of ways in which a necklace can be formed by using 5 identical red beads and 6 identical white beads is:

Detailed Solution for Test: Permutations And Combinations - 1 - Question 25

No of ways to make a necklace out of 11 beads = (11-1)!/2
= 10!/2
Out of 11 beads there are identical red and black heads.
No. of arrangements = (10)!/[2*(5!6!)]

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests
Information about Test: Permutations And Combinations - 1 Page
In this test you can find the Exam questions for Test: Permutations And Combinations - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Permutations And Combinations - 1, EduRev gives you an ample number of Online tests for practice

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests