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Mathematics Exam  >  Mathematics Tests  >  Test:- Permutations And Combinations - 9 - Mathematics MCQ

Test:- Permutations And Combinations - 9 - Mathematics MCQ


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20 Questions MCQ Test - Test:- Permutations And Combinations - 9

Test:- Permutations And Combinations - 9 for Mathematics 2024 is part of Mathematics preparation. The Test:- Permutations And Combinations - 9 questions and answers have been prepared according to the Mathematics exam syllabus.The Test:- Permutations And Combinations - 9 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test:- Permutations And Combinations - 9 below.
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Test:- Permutations And Combinations - 9 - Question 1
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If nCr-1 = 56, nCr = 28 and nCr+1= 8 then r is equal to 

Detailed Solution for Test:- Permutations And Combinations - 9 - Question 1

Solve by simplifying the ratio

Test:- Permutations And Combinations - 9 - Question 2
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 The value of   is equal to

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

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Test:- Permutations And Combinations - 9 - Question 3
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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

Detailed Solution for Test:- Permutations And Combinations - 9 - Question 3

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

Test:- Permutations And Combinations - 9 - Question 4
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 The value of  is:
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 4


∴ 

Test:- Permutations And Combinations - 9 - Question 5
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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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 5

r = 5 & n - r = 8 ⇒ n = 13

Test:- Permutations And Combinations - 9 - Question 6
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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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 6

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

Test:- Permutations And Combinations - 9 - Question 7
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The total number of 9-digit numbers of different digits is :
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 7

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. 

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

 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!

Test:- Permutations And Combinations - 9 - Question 9
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The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 9

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

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

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!

Test:- Permutations And Combinations - 9 - Question 11
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The total number of 5-digit numbers of different digits in which the digit in the middle is the large is 
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 11

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

Test:- Permutations And Combinations - 9 - Question 12
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 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 
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 12

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).

Test:- Permutations And Combinations - 9 - Question 13
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 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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 13

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).

Test:- Permutations And Combinations - 9 - Question 14
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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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 14

'-' 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).

Test:- Permutations And Combinations - 9 - Question 15
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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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 15

Arrange vowels and consonants separately

Test:- Permutations And Combinations - 9 - Question 16
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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
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 16


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.

Test:- Permutations And Combinations - 9 - Question 17
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Let S be the set of all functions from the set A to the set A. If n(A) = k then n(S) is
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 17

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.

Test:- Permutations And Combinations - 9 - Question 18
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 Let A be the set. of 4-digit number a1a2a3a4 where a1>a2>a3>a4, then n(A) is equal to
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 18

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

Test:- Permutations And Combinations - 9 - Question 19
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 The number of numbers divisible by 3 that can be formed by four different even digits is
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 19

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

Test:- Permutations And Combinations - 9 - Question 20
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The number of 5 digit even numbers that can be made with the digits 0, 1, 2 and 3 is
Detailed Solution for Test:- Permutations And Combinations - 9 - Question 20

The last place will be occupied by 0 or 2.

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