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Test:- Permutations And Combinations - 9


20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test:- Permutations And Combinations - 9


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This mock test of Test:- Permutations And Combinations - 9 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Test:- Permutations And Combinations - 9 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test:- Permutations And Combinations - 9 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Test:- Permutations And Combinations - 9 exercise for a better result in the exam. You can find other Test:- Permutations And Combinations - 9 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

If nCr-1 = 56, nCr = 28 and nCr+1= 8 then r is equal to 

Solution:

Solve by simplifying the ratio

QUESTION: 2

 The value of   is equal to

Solution:

QUESTION: 3

In a group of boys, the number of arrangements of 4 boys is 12 times the number of arrangements of 2 boys. The number of boys in the group is

Solution:

Solve by permuting 2 and 4 and making the equation out of it.

QUESTION: 4

 The value of  is:
Solution:


∴ 

QUESTION: 5

From a group of persons, the number of ways of selecting 5 persons is equal to that of 8 persons. The number of persons in the group is
Solution:

r = 5 & n - r = 8 ⇒ n = 13

QUESTION: 6

The number of distinct rational numbers x such that 0<x<1 and  x = P/q,  where p,q ∈{1,2,3,4,5,6}, is
Solution:

As 0 < x < 1, we have p < q.
The number of rational numbers = 5 + 4 + 3 + 2 + 1 = 15
When p, q have a common factor, we get some rational numbers which are not different from those already counted. There are 4 such numbers:

∴ The required number of rational numbers
= 15-4 = 11

QUESTION: 7

The total number of 9-digit numbers of different digits is :
Solution:

The first place from the left can' be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in 9P8 ways.
∴ the number of 9 -digit numbers = 9 x 9P8. 

QUESTION: 8

The number of 6 digit numbers that can be made with the digits 0. 1,2, 3, 4 and 5 so that even digits occupy odd places, is
Solution:

 x | x | x | Crosses can be filled in 3P3 - 2P2, ways
(∴ 0 cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers = (3P3-2P2) x 3!

QUESTION: 9

The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is
Solution:

Consider those 3 oersons as a single unit first, 6! 3!

QUESTION: 10

Seven different lecturers are to deliver lectures in seven periods of a class on a particular day, A, B and C are three of the lecturers. The number of ways in which a routine for the day can be made such that A delivers his lecture before B  and B before C, is
Solution:

As the order of A, B, C is not to change they are to be treated identical in arrangement. So, the required number of ways = 7!/3!

QUESTION: 11

The total number of 5-digit numbers of different digits in which the digit in the middle is the large is 
Solution:

The number of numbers with 4 in the middle = 4P4-3P3
(∴ the other four places are to be filled by 0, 1,2 and 3, and a number cannot begin with 0).
Similarly, the number of numbers with 5 in the middle

QUESTION: 12

 A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done 
Solution:

The numbers are made of 1, 2, 3, 4, 5 or 0, 1 ,2 , 4 , 5.
∴ the required number of 5 digit numbers = 5!+(5P5-4P4).

QUESTION: 13

 Let A a {x | x is prime number and x < 30}. The number of different rational numbers whose numerator and denominator belong to A is
Solution:

A - {2 , 3 , 5, 7 , 11, 13, 17, 19, 23 , 29}. A rational number is made by taking any two in any order.
∴ the required number of rational numbers = 10P2+ 1(including 1).

QUESTION: 14

The total number of ways in which six '+' and four '-' signs can be arrange in a line such that no two ‘-' signs occur together is
Solution:

'-' signs will be put between two '+’ signs or at the two ends.
There arc 7 places for four '-' signs. So, the required number of ways

(there being no arrangement as the '+' signs are identical as well '-' as signs arc identical).

QUESTION: 15

The total number of words that can be made by- writing the letters of the word PARAMETER so that no vowel is between two consonants is
Solution:

Arrange vowels and consonants separately

QUESTION: 16

The number of numbers of four different digits that can be formed from the digits of the number 12 356 such that, the numbers are divisible by 4, is
Solution:


The units place can be filled in 2 ways.
The tens place can be filled in 3 ways.
The remaining two places can be filled by any two of the remaining three digits. So, the required number of numbers = 2 * 3 x 3P2.

QUESTION: 17

Let S be the set of all functions from the set A to the set A. If n(A) = k then n(S) is
Solution:

Each element of the set A can be given the image in the set A in k ways.
∴ the required number of functions,
i.e., n(S) = k x k x ... x (k times)= kk.

QUESTION: 18

 Let A be the set. of 4-digit number a1a2a3a4 where a1>a2>a3>a4, then n(A) is equal to
Solution:

Any selection of four digits from the ten digits 0, 1, 2, 3, ... 9 gives one number. So, the required number of numbers = 10C4.

QUESTION: 19

 The number of numbers divisible by 3 that can be formed by four different even digits is
Solution:

 The numbers will be made by 0, 2 , 4 , 6 or 0 , 4, 6, 8
∴ the required number of numbers 

QUESTION: 20

The number of 5 digit even numbers that can be made with the digits 0, 1, 2 and 3 is
Solution:

The last place will be occupied by 0 or 2.

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