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Permutations and Combinations - 2 - JEE MCQ


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30 Questions MCQ Test - Permutations and Combinations - 2

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


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Permutations and Combinations - 2 - Question 2


Detailed Solution for Permutations and Combinations - 2 - Question 2



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Permutations and Combinations - 2 - Question 3

There are n white and n black balls marked 1, 2,......., n. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours is

Detailed Solution for Permutations and Combinations - 2 - Question 3

Beginning with a white ball, we will get the sequence W, B, W, B,....., containing n white and n black balls. There are n! ways of laying out the W’s, and for each of them, n! ways of arranging the B’s. So the number of arrangements starting with a W is (n!)(n!) = (n!)2 . We can also start with a B, and the nubmer of these arrangements is also (n!)2. Therefore, the total number of arrangements is (n!)2 + (n!)2 = 2(n!)2.

Permutations and Combinations - 2 - Question 4


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Permutations and Combinations - 2 - Question 5


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Permutations and Combinations - 2 - Question 6

f : {1, 2, 3, 4, 5} → {x, y, z, t}. Total number of on to functions ‘f’, is equal to -

Detailed Solution for Permutations and Combinations - 2 - Question 6

Total functions = 45.

    Total functions when any one element is the left out = 4C1. 35

    Total functions when any two elements are left out = 4C2. 25

    Total functions when any two elements are left out = 4C3. 15

    ⇒ Total number of onto functions is equal to 

    45 – (4C1 35 + 4C2. 254C3) = 240

    Ans. (d)


Permutations and Combinations - 2 - Question 7


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Permutations and Combinations - 2 - Question 8


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Permutations and Combinations - 2 - Question 9


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Permutations and Combinations - 2 - Question 10

The number of ways in which a mixed double game can be arranged from amongst 9 married couples if no husband and wife play in the same game is -

Detailed Solution for Permutations and Combinations - 2 - Question 10


Permutations and Combinations - 2 - Question 11


Detailed Solution for Permutations and Combinations - 2 - Question 11


Permutations and Combinations - 2 - Question 12

An old man while dialing a 7 digit telephone number remembers that the first four digits consists of one 1's, one 2's and two 3's. He also remembers that the fifth digit is either a 4 or 5 while has no memorising of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number of distinct trials he has to try to make sure that he dials the correct telephone number, is

Detailed Solution for Permutations and Combinations - 2 - Question 12


Permutations and Combinations - 2 - Question 13


Detailed Solution for Permutations and Combinations - 2 - Question 13


Permutations and Combinations - 2 - Question 14


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Permutations and Combinations - 2 - Question 15


Detailed Solution for Permutations and Combinations - 2 - Question 15


Permutations and Combinations - 2 - Question 16


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Permutations and Combinations - 2 - Question 17

In a jeep there are 3 seats in front and 3 in the back. Number of different ways can six persons of different heights be seated in the jeep, so that every one in front is shorter than the person directly behind is

Detailed Solution for Permutations and Combinations - 2 - Question 17


Permutations and Combinations - 2 - Question 18


Detailed Solution for Permutations and Combinations - 2 - Question 18


Permutations and Combinations - 2 - Question 19

Number of 7 digit numbers the sum of whose digits is 61 is

Detailed Solution for Permutations and Combinations - 2 - Question 19


Permutations and Combinations - 2 - Question 20

A four digit number is called a doublet if any of its digit is the same as only one neighbour.  For example, 1221 is a doublet but 1222 is not. Find the number of such doublets.

Detailed Solution for Permutations and Combinations - 2 - Question 20

Case-1:    abbc        

 We can choose a in 9 ways, b in 9 ways, and c in 9 ways. Hence, there are 93 possibilities for this case

Case-2:    bbac

There are 9 possibilities for b, and 9 possibilities for each a and c. Once again, we have that there are 93 possibilities for this case

Case-3:    acbb

Once again, there are 9 possibilities for a and c, and 9 possibilities for b. Hence , there are 93 possibilities for this case

Case-4:    aabb

We have 9 possibilities for a and 9 possibilities for b. We thus have 81 possibilities for this case.

We can now sum up our answers, to get a total of  3 · 93 + 81 = 2268 total ways  Ans

Permutations and Combinations - 2 - Question 21

Let  Pn  denotes the number of ways of selecting 3 people out of  ' n '  sitting in a row ,  if no two of them are consecutive and  Qn  is the corresponding figure when they are in a circle . If   Pn - Qn  =  6 ,  then  ' n '  is equal to :

Detailed Solution for Permutations and Combinations - 2 - Question 21


Permutations and Combinations - 2 - Question 22


Detailed Solution for Permutations and Combinations - 2 - Question 22


Permutations and Combinations - 2 - Question 23

Number of positive integers which have no two digits having the same value with sum of their digits being 45, is

Detailed Solution for Permutations and Combinations - 2 - Question 23


Permutations and Combinations - 2 - Question 24

Number of 4 digit numbers of the form  N = abcd  which satisfy following three conditions    

(i)    4000 ≤ N < 6000    

(ii)    N is a multiple of 5        

(iii)    3 ≤ b < c ≤ 6 is equal to

Detailed Solution for Permutations and Combinations - 2 - Question 24


Permutations and Combinations - 2 - Question 25

The number of positive integral solutions of the equation x1 x2 x3 x4 x5 = 1050, is

Detailed Solution for Permutations and Combinations - 2 - Question 25


Permutations and Combinations - 2 - Question 26

In a conference 10 speakers are present. If  S1 wants to speak before S2 & S2 wants to speak after S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is

Detailed Solution for Permutations and Combinations - 2 - Question 26


Permutations and Combinations - 2 - Question 27


Detailed Solution for Permutations and Combinations - 2 - Question 27


Permutations and Combinations - 2 - Question 28

Find the number of  4 digit numbers starting with 1 and having exactly two identical digits.

Detailed Solution for Permutations and Combinations - 2 - Question 28


Permutations and Combinations - 2 - Question 29


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Permutations and Combinations - 2 - Question 30

A crew of an eight oar boat has to be chosen out of 11 men five of whom can row on stroke side only, four on the bow side only, and the remaining two on either side. Find the different selections that can be made.

Detailed Solution for Permutations and Combinations - 2 - Question 30


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