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Permutations and Combinations - 1 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Permutations and Combinations - 1

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

Permutations and Combinations - 1 - Question 2

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 Permutations and Combinations - 1 - Question 2

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

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Permutations and Combinations - 1 - Question 3

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

Detailed Solution for Permutations and Combinations - 1 - Question 3

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

Permutations and Combinations - 1 - Question 4

If nP5 = 60n−1P3, then n is

Detailed Solution for Permutations and Combinations - 1 - Question 4

 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.

Permutations and Combinations - 1 - Question 5

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

Detailed Solution for Permutations and Combinations - 1 - Question 5

 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

Permutations and Combinations - 1 - Question 6

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 Permutations and Combinations - 1 - Question 6

To solve this problem, we need to calculate the total number of ways a candidate can attempt the multiple-choice question where there are 4 alternatives, and one or more can be correct.

Step 1: Consider each alternative

Each of the 4 alternatives can either be:

  1. Selected (included in the answer)
  2. Not selected (excluded from the answer)

So, for each alternative, there are 2 choices (select or not select).

Step 2: Calculate the total number of combinations

If there were no restrictions (i.e., selecting none is allowed), the total number of combinations would be 24=16

Step 3: Subtract the invalid case

Since at least one alternative must be selected (one or more are correct), we subtract the one case where none of the alternatives are selected:

24−1=16−1=15

Permutations and Combinations - 1 - Question 7

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

Permutations and Combinations - 1 - Question 8

All possible  three digits even numbers which can be formed with the condition that if 5 is one of the digit, then 7 is the next digit is :

Detailed Solution for Permutations and Combinations - 1 - Question 8


Permutations and Combinations - 1 - Question 9

The number of ways in which we can distribute mn students equally among m sections is given by-

Detailed Solution for Permutations and Combinations - 1 - Question 9


Permutations and Combinations - 1 - Question 10


Detailed Solution for Permutations and Combinations - 1 - Question 10


Permutations and Combinations - 1 - Question 11


Detailed Solution for Permutations and Combinations - 1 - Question 11


Permutations and Combinations - 1 - Question 12


Detailed Solution for Permutations and Combinations - 1 - Question 12


Permutations and Combinations - 1 - Question 13

Find the number of five digit numbers that can be formed using the digits   1, 2, 3, 4, 5, 6, 7, 8, 9   in which one digit appears once and two digits appear twice (e.g.  41174  is one such number but  75355  is not.).

Detailed Solution for Permutations and Combinations - 1 - Question 13


Permutations and Combinations - 1 - Question 14

A set contains 2n + 1 elements. The number of subsets of this set which contains more than n elements is -

Detailed Solution for Permutations and Combinations - 1 - Question 14


Permutations and Combinations - 1 - Question 15


Detailed Solution for Permutations and Combinations - 1 - Question 15


Permutations and Combinations - 1 - Question 16

Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of them gets atleast two blankets, equal to

Detailed Solution for Permutations and Combinations - 1 - Question 16


Permutations and Combinations - 1 - Question 17


Detailed Solution for Permutations and Combinations - 1 - Question 17


Permutations and Combinations - 1 - Question 18


Detailed Solution for Permutations and Combinations - 1 - Question 18


Permutations and Combinations - 1 - Question 19

There are counters available in x different colours. The counters are all alike except for the colour. The total number of arrangements consisting of y counters, assuming sufficient number of counters of each colour, if no arrangement consists of all counters of the same colour is :

Detailed Solution for Permutations and Combinations - 1 - Question 19

 xy - x

Permutations and Combinations - 1 - Question 20


Detailed Solution for Permutations and Combinations - 1 - Question 20


Permutations and Combinations - 1 - Question 21

The total number of six digit numbers x1 x2 x3 x4 x5 x6 having the property that x1 < x2 ≤ x3 < x4 < x5 ≤ x6 is equal to -

Detailed Solution for Permutations and Combinations - 1 - Question 21


Permutations and Combinations - 1 - Question 22


Detailed Solution for Permutations and Combinations - 1 - Question 22


Permutations and Combinations - 1 - Question 23

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. If internal arrangement inside the car does not matter then the number of ways in which they can travel, is

Detailed Solution for Permutations and Combinations - 1 - Question 23


Permutations and Combinations - 1 - Question 24

 A committee of 5 is to be chosen from a group of 9 people. Number of ways in which it can be formed if two particular persons either serve together or not at all and two other particular persons refuse to serve with each other, is

Detailed Solution for Permutations and Combinations - 1 - Question 24


Permutations and Combinations - 1 - Question 25


Detailed Solution for Permutations and Combinations - 1 - Question 25


Permutations and Combinations - 1 - Question 26


Detailed Solution for Permutations and Combinations - 1 - Question 26


Permutations and Combinations - 1 - Question 27


Detailed Solution for Permutations and Combinations - 1 - Question 27

Permutations and Combinations - 1 - Question 28


Detailed Solution for Permutations and Combinations - 1 - Question 28


Permutations and Combinations - 1 - Question 29

 Let Pn  denotes the number of ways in which three people can be selected out of  ' n '  people sitting in a row ,  if no two of them are consecutive.  If    Pn + 1  -  Pn  =  15  then the value of  'n'  is

Detailed Solution for Permutations and Combinations - 1 - Question 29


Permutations and Combinations - 1 - Question 30

4 letter lock consists of three rings each marked with 10 different letters, the number of ways in which it is possible to make an unsuccessful attempt to open the lock, is

Detailed Solution for Permutations and Combinations - 1 - Question 30


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