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Test:- Permutations And Combinations - 5 - Mathematics MCQ


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test:- Permutations And Combinations - 5

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Test:- Permutations And Combinations - 5 - Question 1

In how many ways 100 runs can be scored with the help of four's and six's

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 1

25 four and 0 sixes and so on there are 9 cases

Test:- Permutations And Combinations - 5 - Question 2

In a certain test there arc n questions. In this test, 2n-k student gave wrong answers to at least k questions, where k 1,2,3,4..., n. If the total number of wrong answers given is 2047, then n is equal to :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 2

2n- 1 wrong answer are there 
2n - 1 = 2047
⇒ n = 11

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Test:- Permutations And Combinations - 5 - Question 3

The total number of 7 digits numbers the sum of whose digits is even is :

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

Total 7 digit no. = 9x106
sum will be even in half cases
9000000/2 = 4500000

Test:- Permutations And Combinations - 5 - Question 4

To fill 12 vacancies there are 25 candidates of which 5 are from scheduled casts. If 3 of the vacancies are reserved for scheduled casts candidates while the rest are open to all. The number of ways in which the selections can be made is :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 4

Total vacancies = 12
total candidate = 25
number of scheduled caste = 5 out of 25.
reserved vacancies = 3
total ways of selection = 5C3 x 22C9

Test:- Permutations And Combinations - 5 - Question 5

A businessman hosts a dinner to 21 guests. He is having 2 round tables which can accommondate 15 and 6 persons. How many ways are there to host dinner?

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

Total guest = 21, guests are
seated in round table respectively are 15 and 6 
Total ways = 21C15 x 14!5!

Test:- Permutations And Combinations - 5 - Question 6

In how many ways can 8 men will be seated around a round table when in no two ways a man has the same neighbors?

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 6

8 men seated around a round table
with no same neighbour

Test:- Permutations And Combinations - 5 - Question 7

A question paper consists of two sections having respectively 3 and 4 questions. The following note is given on the paper. “It is not necessary to attempt all the questions one question from each section is compulsory'. “In how' many ways the candidate can select the 5 questions?

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 7

 Total ways = 7C3 = 21

Test:- Permutations And Combinations - 5 - Question 8

 A candidate is required to answer 7 out of 15 question which are divided into three groups each containing 4,5,6 question respectively. He is required to select at least 2 questions from each group. In how many ways can he make up his choice?

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 8

Candidate required to answer 7 out of  15 15C7

Test:- Permutations And Combinations - 5 - Question 9

 A letter lock consists of three rings each marked with 12 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock.

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

Total number of unsuccessful attempt
= (12)3 - 1
= 1728 - 1 =1727

Test:- Permutations And Combinations - 5 - Question 10

The number of ways of choosing a committee of 4 women and 5 men from 10 women and 9 men, If Mr. A refuse to serve on committee, if Mr. B is a member of the committee, can not exeed.

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 10

 Total women = 10
total men = 9
choosing committee = 4 w and 3 m

Test:- Permutations And Combinations - 5 - Question 11

In a football championship, 153 matches were played. Every team played one match with each other. The number of teams participating in the championship is:

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 11

Test:- Permutations And Combinations - 5 - Question 12

The number of ways in which 52 cards can be divided into 4 set, three of them having 17 cards each and the fourth one having just one cards is:

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 12

52 are divided into 4 sets

Test:- Permutations And Combinations - 5 - Question 13

In how many ways can the letters of the word ‘ARRANGE’ be arranged so that the two A's are together but not two R’s is-

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 13

 A R R A N G E

Test:- Permutations And Combinations - 5 - Question 14

In an examination there are three multiple questions and each question has 4 choice. Number of ways in which a student can fail to get all answer correct is-

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 14

Total 3 multiple question each multiple question has 4 choice
number of ways for student to fail to get all answer correct 
43-1 =64-1= 63

Test:- Permutations And Combinations - 5 - Question 15

 There are three straight lines which are parallel and lie in the same plane A total number of 5 point on the first, 7 points on the second, mid 8 points on the third straight line as selected. The maximum number of triangles formed with the vertices at these points are :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 15

 Total triangle formed = 

Test:- Permutations And Combinations - 5 - Question 16

The number of ways in which a team of eleven players can be selected from 22 players including 2 of them and excluding 4 of them is:

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 16

Number of way 11 player will be selected

Test:- Permutations And Combinations - 5 - Question 17

Seven women and seven men are to sit round a circular table such that there is a man on either side of every women; the number of seating arrangements is :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 17

 6! x 7!

Test:- Permutations And Combinations - 5 - Question 18

The number of ways in which four letters can be selected from the word ‘RAMANEA’ is 

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 18

R A M A N E A has R1, M1, N2 E - l A -3
Total ways = 

Test:- Permutations And Combinations - 5 - Question 19

The number of ways in which seven persons can be arranged at a round table if two particular persons may not seat together is :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 19

 6! - 2!5!

= 720 - 240

= 480

Test:- Permutations And Combinations - 5 - Question 20

The number of ways in which 4 boys and 4 girls can form a line, with boys and girls alternating, is :

Detailed Solution for Test:- Permutations And Combinations - 5 - Question 20

B G B G B O B G
boys can be seated in 4! ways
girls can be seated in 4! ways
But Boy & girls are seated into 2 ways
So, 2(4!)2 is total ways

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