Test:- Permutations And Combinations - 2


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


Description
This mock test of Test:- Permutations And Combinations - 2 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Test:- Permutations And Combinations - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test:- Permutations And Combinations - 2 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Test:- Permutations And Combinations - 2 exercise for a better result in the exam. You can find other Test:- Permutations And Combinations - 2 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
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?

Solution:

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)

QUESTION: 2

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

Solution:

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

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

Solution:

Total handshakes = nC2 = 66


⇒ n = 12 (only) as negative answer is not valid.

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 :

Solution:

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

QUESTION: 5

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

Solution:



⇒  n - 9 and r = 3

QUESTION: 6

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

Solution:


QUESTION: 7

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

Solution:




so, r = 14

QUESTION: 8

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

Solution:


QUESTION: 9

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

Solution:

QUESTION: 10

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

Solution:


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:

Solution:

BROTHER   B, E, T, H, O : 1, R : 2

Total = 249

QUESTION: 12

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

Solution:

QUESTION: 13

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

Solution:

The sum of all number formed by the digit 1,3,5,7 is

QUESTION: 14

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

Solution:

The total number of distinct function is 1010.

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 :

Solution:

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.

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 :

Solution:

QUESTION: 17

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

Solution:

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

QUESTION: 18

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

Solution:

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!

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 :

Solution:

Last place we will be of 5 types because we exclude 5 there and a fter that simply use all freely.

QUESTION: 20

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

Solution:

We have a very good formula

where x in number of diagonal and n stand for number of sides so putting the value.