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Test:- Permutations And Combinations - 3 - Mathematics MCQ


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20 Questions MCQ Test - Test:- Permutations And Combinations - 3

Test:- Permutations And Combinations - 3 for Mathematics 2024 is part of Mathematics preparation. The Test:- Permutations And Combinations - 3 questions and answers have been prepared according to the Mathematics exam syllabus.The Test:- Permutations And Combinations - 3 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test:- Permutations And Combinations - 3 below.
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Test:- Permutations And Combinations - 3 - Question 1

A parallelogram is cut by two sets of m lines parallel to its sides. The number to parallelogram thus formed is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 1

We have the total number of horizontal lines as m + 2 and verticle lines as m + 2 also.
and we take 2 line from both side to make a parallelogram so number of ways = 

Test:- Permutations And Combinations - 3 - Question 2

The number of paralleograms that can be formed from a set of seven parallel lines intersecting another set of seven parallel lines is :

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

The number of paralleogram with the help of 7 parallel and 7 horizontal line =  (21)2 = 441

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Test:- Permutations And Combinations - 3 - Question 3

The number of triangles that are formed by choosing the vertices from a set of 12 point, seven of which lies on the same straight line is :

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

We have points where 7 point are in line and other 5 points are non-coliinear so the triangles = 

(As non-collinear points form triangle)

Test:- Permutations And Combinations - 3 - Question 4

The number of positive integral solutions of abc = 42

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 4

Test:- Permutations And Combinations - 3 - Question 5

The number of positive integral solutions of abed = 210

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 5


⇒ 

Test:- Permutations And Combinations - 3 - Question 6

The number of triangles whose vertices are at the vertices of an octagon but none of whose side is the side of the octagon is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 6

Total number of triangle - Number of triangle having one side common - number of triangle having two side common with octagon.

Test:- Permutations And Combinations - 3 - Question 7

Three dice are rolled. The number of possible outcomes in which at least one die shows 5 is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 7

All possibility in which all dice show’ values.
= 6 x 6 x 6
All possibility in which no dice shows 5
= 5 x 5 x 5
- 216 - 125 = 91

Test:- Permutations And Combinations - 3 - Question 8

How many words can be formed from the letters of the word ‘DAUGHTER’ so that the vowels come together?

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 8

We have 5 consonants and 3 vowels first of all we treat 3 vowels as one letter the we arrange them in 6! way after that we arrange 3 vowels in 3 ways so total ways
= 6! x 3!
= 720 x 6 = 4320

Test:- Permutations And Combinations - 3 - Question 9

In how many ways three girls and nine boys can be seated in two vans, each having numbered seat, 3 in the front and 4 at the back?

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

There is no boundation so we can take
14P12 as number pf ways
= 14! / 2! = 14 x 13 x 12 x 11 x 10 x ... x 3
 

Test:- Permutations And Combinations - 3 - Question 10

The greatest possible number of points of intersection of 8 straight lines and 4 circle is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 10

When we take two circle then they can cut maximum in two point when two lines are taken then 1 point of and one circle and one line then 2 point of so total number.

Test:- Permutations And Combinations - 3 - Question 11

The number of ways of painting the faces of a cube with six different colour is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 11

We have one side of cube to point at a time so total the way will be single (1).

Test:- Permutations And Combinations - 3 - Question 12

The number of ways in which 8 different flowers can be stung to from a garland so that 4 particular flowers are never separated is:

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 12

The total number of way

Test:- Permutations And Combinations - 3 - Question 13

The number of all possible selection which a student can make for answering one or more questions out of eight given question in paper, when each question has an alternative is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 13

There are three options for 1 question for a student (i) either he attempt main part (ii) or part (iii) or none of main or or part so every question can be handled in three ways.
So, total ways

-1 so that he has to attempt necessarily one question.

Test:- Permutations And Combinations - 3 - Question 14

The sum of all number greater than 1000 formed by using digits 1,3,5,7, no digit being repeated in any number is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 14

With the help of digit 1, 3, 5, 7 all together ail number are greater than 1000 so, total sum.
= (1 + 3 + 5 + 7) x 1111 x 3!
16 x 1111 x 6
= 106656

Test:- Permutations And Combinations - 3 - Question 15

A father with eight children takes 3 at a time to the circus, as often as he can without taking the same 3 children together more than once, the number of times he can go to circus is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 15

The total number to visit circus is 8c3 = 56

Test:- Permutations And Combinations - 3 - Question 16

The number of ways in which four letters can be selected from the word KAMANEA is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 16

Write all the varieties before evaluation

Test:- Permutations And Combinations - 3 - Question 17

The number of ways in which the letters of the word ‘RAMANEA’ are arranged such that, the relative position of vowels and consonant are not changed is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 17

Relative position means every word should maintain it’s order

Total ways = 

Test:- Permutations And Combinations - 3 - Question 18

If in class tournament each class plays once against, each of the other and in all 45 games are played, then the number of participants is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 18


Test:- Permutations And Combinations - 3 - Question 19

If 56Pr+6 : 56Pr+3 = 1320:1 Then the value of r is:

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 19


Test:- Permutations And Combinations - 3 - Question 20

A person goes in for examination in which there are four papers with a maximum of 20 marks from each paper. The number of ways in which one can get 40 marks is :

Detailed Solution for Test:- Permutations And Combinations - 3 - Question 20

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