Test:- Permutations And Combinations - 2

# Test:- Permutations And Combinations - 2 - Mathematics

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## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics - Test:- Permutations And Combinations - 2

Test:- Permutations And Combinations - 2 for Mathematics 2023 is part of Topic-wise Tests & Solved Examples for IIT JAM Mathematics preparation. The Test:- Permutations And Combinations - 2 questions and answers have been prepared according to the Mathematics exam syllabus.The Test:- Permutations And Combinations - 2 MCQs are made for Mathematics 2023 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

### How many numbers divisible by 5 and lying between 3000 and 5000 can be formed from the digit 0,3,4,5, and 7?

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

First place will be filled by 3 or 4 2nd by 5, 3rd & 4th by 2 so total way.
= 2 x 5 x 5 x 2 = 100-1
= 99 ; (-1 is to exclude 3000)

Test:- Permutations And Combinations - 2 - Question 2

### Total number of natural numbers less than 4000 formed with digits 0,1,2,3,4,5 and 8 is :

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

The one digit number will be 6
The 2 digit number will be
6 x 7 = 42
The 3 digit number will be
6 x 7 x 7 = 294
The 4 digit unit number will be
3 x 7 x 7 x 7 = 1029
so total number from 0 to 4000 excluding 4000 will be
6 + 42 + 294+ 1029= 1371

Test:- Permutations And Combinations - 2 - Question 3

### Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of person in the room is

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

Total handshakes = nC2 = 66  ⇒ n = 12 (only) as negative answer is not valid.

Test:- Permutations And Combinations - 2 - Question 4

The sum of the digits in the digits unit place of all the numbers formed with the help of 3,4,5,6 taken all at a time is :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 4

We can arrange this number in 4! way i.c. 24 way and each digit appear 3! times at each place, so each number come on unit place 6 time so sum = (3 + 4 + 5 +6) * 6
= 18  x 6 = 108

Test:- Permutations And Combinations - 2 - Question 5

If nCn-1 =  36; nCr = 84 and nCr+1 = 126, then r is equal to:

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 5  ⇒  n - 9 and r = 3

Test:- Permutations And Combinations - 2 - Question 6

If nPr= nPr+1 and nCr= nCr-1 then (n,r) are

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 6  Test:- Permutations And Combinations - 2 - Question 7

If 28Cr : 24Cn+4 = 225 : 11, then

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 7   so, r = 14

Test:- Permutations And Combinations - 2 - Question 8

If nCr + nCr+1 = n+3Cx, then x =

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 8  Test:- Permutations And Combinations - 2 - Question 9

The total number of arrangement that can be made out of the letter of the word RAMANEA is:

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 9 Test:- Permutations And Combinations - 2 - Question 10

If n-1c3 / n-1c4 > nc3, then the least value of n is:

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 10  Test:- Permutations And Combinations - 2 - Question 11

If the letter of the word BROTHER are written in all possible order and these word are written out as in a dictionary, then the rank of the word BROTHER is:

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

BROTHER   B, E, T, H, O : 1, R : 2 Total = 249

Test:- Permutations And Combinations - 2 - Question 12

The rank of the word ‘RAMANEA’ in the dictionary made by the letters of this word is:

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 12 Test:- Permutations And Combinations - 2 - Question 13

The sum of the numbers formed by the digit 1,3,5 and 7 taking all at a time is:

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 13

The sum of all number formed by the digit 1,3,5,7 is Test:- Permutations And Combinations - 2 - Question 14

Let A be a set containing 10 distinct elements, Then the total number of distinct function from A to A is. :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 14

The total number of distinct function is 1010.

Test:- Permutations And Combinations - 2 - Question 15

There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is :

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

The total number of way in which way we can do switches are
210 -1 = 1024-1=1023
-1 so that at least one switch will be on for light in the hall.

Test:- Permutations And Combinations - 2 - Question 16

Out of 16 players of cricket team, 4 are bowlers and 2 are wicket keepers. A team of 11 players is to be chosen so as to contain at least 3 bowlers and at least one wicket keeper. The number of ways in which the team be selected is :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 16 Test:- Permutations And Combinations - 2 - Question 17

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 17

Selecting 2 vowels and 3 consonants from 4 vowels and 5 consonants will be in and the number of way in which we can arrange the 5 letters is 5!
so, 60 x 5! = 60 x 120 = 7200

Test:- Permutations And Combinations - 2 - Question 18

The total number of 9 digits number which have all different digits is :

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

We have number from 0 to 9 so on first place we will put any 9 letter and after that we put digits into remaining place in 9P8 = 9! ways
so, total ways
= 9x 9!

Test:- Permutations And Combinations - 2 - Question 19

The number of different number of six digits each (without repetition of digit) can be formed from the digits 4,5,6.7,8,9 such that they are not divisible by 5 is :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 19

Last place we will be of 5 types because we exclude 5 there and a fter that simply use all freely. Test:- Permutations And Combinations - 2 - Question 20

A polygon has 54 diagonals, then the number of its sides arc :

Detailed Solution for Test:- Permutations And Combinations - 2 - Question 20

We have a very good formula where x in number of diagonal and n stand for number of sides so putting the value. ## Topic-wise Tests & Solved Examples for IIT JAM Mathematics

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