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Test: Permutations Non Distinct Objects - JEE MCQ


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15 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Permutations Non Distinct Objects

Test: Permutations Non Distinct Objects for JEE 2025 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Permutations Non Distinct Objects questions and answers have been prepared according to the JEE exam syllabus.The Test: Permutations Non Distinct Objects MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutations Non Distinct Objects below.
Solutions of Test: Permutations Non Distinct Objects questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Permutations Non Distinct Objects 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 Non Distinct Objects | 20 questions in 20 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 Non Distinct Objects - Question 1
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Sonia has 10 balloons out of which 5 are red, 2 white, 2 blue and 1pink, which she wants to use for the decoration. Her favourite pink colour balloon should be filled with toffees and should be put at the centre of the room above the cake table and remaining 9 at the wall behind the cake table. How many ways she can arrange the balloons?

Detailed Solution for Test: Permutations Non Distinct Objects - Question 1

1 balloon is fixed that is to be placed above the cake
Now she have to arrange 9 balloons = 9!/(5!*2!*2!)
(9*8*7*6*5!)/((5!*2*2)
= (9*8*7*6)/(2*2)
= 756

Test: Permutations Non Distinct Objects - Question 2
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In how many ways can 3 letters be posted in 4 letter boxes?

Detailed Solution for Test: Permutations Non Distinct Objects - Question 2

The first letter can be posted in 4 ways. So, total outcomes about the first letter = 4.
For every outcome about the first letter, the second letter can be posted in 4 ways. So, total outcomes about the first and the second letters= (4*4) = 16.
Therefore, following the same route, we can say, total possible outcomes about the first and the second letters= (4*4*4) = 64.

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Test: Permutations Non Distinct Objects - Question 3
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Among 7 flags 4 are of red colour and the rest are all different colours. How mant different signals can be generated using these flags?

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Test: Permutations Non Distinct Objects - Question 4
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 If (n + 1)! = 20(n – 1)!, then n is equal to
Detailed Solution for Test: Permutations Non Distinct Objects - Question 4

(n + 1)! = 20 (n – 1)!
n (n + 1) = 20
(n – 4) (n + 5) = 0          
Since, (n – 1)! exists, n ≥ 1
So, n = 4 

Test: Permutations Non Distinct Objects - Question 5
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There are 3 white, 4 red and 1 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all 8 marbles are drawn, determine the number of different arrangements if marbles of same colour are indistinguishable.
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Test: Permutations Non Distinct Objects - Question 6
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The number of permutations of n different objects taken r at a time, where repetition is allowed
Detailed Solution for Test: Permutations Non Distinct Objects - Question 6

Number of permutations of n different things taken r at a time when repetition is allowed = nr

Test: Permutations Non Distinct Objects - Question 7
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In a class 6 students have to be arranged for a photograph. If the prefect and the vice must occupy the positions at either ends, how many ways the students can be arranged?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 7

In a class of 6 students. Two students at both ends. Remaining four students have to be arranged 
= 2 * 4! 
= 2 * 24
= 48

Test: Permutations Non Distinct Objects - Question 8
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How many different words can be formed using the letters of the word BHARAT, which begin with B and end with T?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 8

If the first & last letter is fixed, then we find out, the number of permutations of the remaining letters, i.e. 4
= 4!/2!
= 4*3*2!/2!
= 12

Test: Permutations Non Distinct Objects - Question 9
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Find the number of words with or without meaning which can be made using all the letters of the word SWEET.
Detailed Solution for Test: Permutations Non Distinct Objects - Question 9

Sweet has 5 letters, so it gives 5! but there are two letters which are same
Number of words will be = (5! ÷ 2!) = 60

Test: Permutations Non Distinct Objects - Question 10
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If n = 4 and r = 2, the value of 
Detailed Solution for Test: Permutations Non Distinct Objects - Question 10

n!/(n-r)!
4!/(4-2)!  = 4*3*2!/2!
= 12

Test: Permutations Non Distinct Objects - Question 11
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Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 11

1/4! + 1/5!
= 1/4![1 + 1/5]
1/4![6/5]
=> 6/5!

Test: Permutations Non Distinct Objects - Question 12
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In how many ways can 10 people line up at a ticket window of a cinema hall?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 12

 If you imagine there are 10 spots on the floor in a line where people can stand, then the first person has 10 choices of where to stand.
The second person has 9 choices since the first person is already on one of the spots.
The third person has 8 choices. And so on until the last person has no choice.
We can now mulitply the numbers together. 10x9x8x7x6x5x4x3x2x1. This is called a factorial (where you multiply all of the consecutive numbers together from 1) and it is written as 10!.
This means there are 3,628,800 ways of arranging the 10 people

Test: Permutations Non Distinct Objects - Question 13
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In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
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Test: Permutations Non Distinct Objects - Question 14
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A coin is tossed 6 times, in how many throws can 4 heads and 2 tails be obtained?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 14

To begin with, we calculate the total number of possibilities that arise from tossing a coin 6 times . On each toss , we have 2 possibilities - a head or a tail. This gives us
2*2*2*2*2*2 = 64 possibilities
Now let's list out the desirable outcomes.
   Having 4 heads
H H H H T T - This is one example of the above outcome. Something like
H H H T H T would also be equally likely and would be a desirable outcome. Thus to calculate all such permutations
= (6!/4!) ∗ 2!
= 15ways

Test: Permutations Non Distinct Objects - Question 15
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How many three letter codes can be formed by only vowels of English alphabets, given that repetition of letters is allowed?
Detailed Solution for Test: Permutations Non Distinct Objects - Question 15

We have 5 choices (a, e, i, o, u)
3 letter codes = 5 * 5 * 5
= 125

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