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Test: Basic Concepts Of Permutations And Combinations- 4 - CA Foundation MCQ


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30 Questions MCQ Test - Test: Basic Concepts Of Permutations And Combinations- 4

Test: Basic Concepts Of Permutations And Combinations- 4 for CA Foundation 2024 is part of CA Foundation preparation. The Test: Basic Concepts Of Permutations And Combinations- 4 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Basic Concepts Of Permutations And Combinations- 4 MCQs are made for CA Foundation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts Of Permutations And Combinations- 4 below.
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Test: Basic Concepts Of Permutations And Combinations- 4 - Question 1

If nC6 ÷ n-2C3 = 91/4 then the value of n is _____________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 2

In forming a committee of 5 out of 5 males and 6 females how many choices you have to made so that there are 3 males and 2 females?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 2

5C3 x 6C2 = 5 x 4/2 x 6 x 5/2

=150

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Test: Basic Concepts Of Permutations And Combinations- 4 - Question 3

 In order to pass PE-II examination minimum marks have to be secured in each of 7 subjects. In how many ways can pupil fail?

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

Given that for a student to pass an examination , a minimum has to be secured in each of the 7 subjects.
A student would fail if he doesn't secure a minimum in any one o
the subjects A student would fail either if he fails in 1 subject or he fails
in 2 subjects or 3 subjects or so on continuing.........or fails in all the
subjects
A student can fail in 1 subject in ⁷C₁ ways
A student can fail in 2 subjects in ⁷C₂ ways
A student can fail in 3 subjects in ⁷C₃ ways
A student can fail in all subjects in ⁷C₇ = 1 way
So, total number of ways in which student would fail an examination are
(⁷C₁ + ⁷C₂ + .......+ ⁷C₇) = 2⁷ - 1 = 127

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 4

 A team of 12 men is to be formed out of n persons. It is found that ‘A’ and ‘B’ are three times as often together as ‘C’ ‘D’ and ‘E’ are. Then the value of n is __________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 5

 You are selecting a cricket team of first 11 players out of 16 including 4 bowlers and 2 wicket keepers. In how many ways you can do it so that the team contains exactly 3 bowlers and 1 wicket keeper?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 6

Out of 10 consonants and 4 vowels how many words can be formed each containing 6 consonant and 3 vowels?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 7

In your college Union Election you have to choose candidates. Out of 5 candidates 3 are to be elected and you are entitled to vote for any number of candidates but not exceeding the number to be elected. You can do it in ________ ways.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 8

 A team of 12 men is to be formed out of n persons. Then the number of times 2 men ‘A’ and ‘B’ are together is ________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 9

In paper from 2 groups of 5 questions each you have to answer any 6 questions attempting at least 2 question from each group. This is possible in ________ number of ways.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 10

 The number of combinations that can be made by taking 4 letters of the word 'COMBINATION'

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 11

 Out of 8 different balls taken three at a time without taking the same three together more than once for how many number of times you can select a particular balls?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 12

 A boat’s crew consist of 8 men, 3 of whom can row only on one side and 2 only on the other. The number of ways in which the crew can be arranged is _______.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 13

Out of 8 different balls taken three at a time without taking the same three together more than once for for how many number of times you can select any ball?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 14

You are selecting a cricket team of first 11 players out of 16 including 4 bowlers and 2 wicket keepers. In how many ways you can do it so that the team contains exactly 3 bowlers and 1 wicket keeper? Would your answer be different if the team contains at least 3 bowlers and at least 1 wicket-keeper?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 14

ANSWER :- A

Solution :- 

There are : 4 bowlers,  2 wicket-keepers, 10 non-bowlers-not-wicket keepers.

There are exactly 3 bowlers, exactly 1 wicket keeper, and exactly 7 non-bowlers-non-wicket-keepers.

(4C3)(2C1)(10C7) 

= 960

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 15

 If 18C18Cn+2 then the value of n is _________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 16

 In your office 4 posts have fallen vacant. In how many ways a selection out of 31 candidates can be made if one candidate is always included?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 16

since 1 candidate is already selected so there are 3 posts for which we need candidates from 30.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 17

A party of 6 is to be formed form 10 men an 7 women so as to include 3 men and 3 women. In how many ways the party can be formed if two particular women refuse to join it?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 18

A team of 12 men is to be formed out of n person. The number of times 3 men ‘C’ ‘D’ and ‘E’ are together is_________.

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 18

Put C,D and E as one unit
∴ Total people to be selected = 10 = 9 ppl + one group of 3
Similarly total no ppl= n-3+1 = n-(C,D,E) + a group of 3 ppl [Note:Here 3 ppl are replaced by one group of 3 ppl ∴ there are counted as one and not 3]

Selection:n-3+1 C10 = n-2C10

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 19

In how many ways can 8 boys form a ring?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 20

The total number of sitting arrangements of 7 persons in a row if 3 persons sit together in a particular order is _________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 21

 In how many ways 6 men can sit at a round table so that all shall not have the same neighbors in any two occasions?

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 22

 From 6 boys and 4 girls 5 are to be seated. If there must be exactly 2 girls the number of ways of selection is __________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 23

If you have to make a choice of 7 questions out of 10 questions set, you can do it in ________ number of ways.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 24

 The total number of sitting arrangements of 7 persons in a row if one person occupies the middle seat is _________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 25

 In how many ways 6 men and 6 women sit at a round table so that no two men are together?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 25

First, we sit the men in an order that there is a place vacant between two men. Then, number of ways sitting 6 men = (6 – 1)! = 5!

Now a number of ways sitting 6 women in the vacant places = 6!

Number of permutations = (6 – 1)! = 5!

Now, the number of ways sitting 6 women in the vacant place = 6!

Number of permutaions = 5! × 6! 

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 26

If all the permutations of the letters of the word “chalk” are written in a dictionary the rank of this word will be __________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 27

In how many ways 4 men and 3 women are arranged at a round table if the women always sit tohether?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 27

If all three women always sit together

We can club the 3 women in one group and the number of arrangements of this group will be 3!

Now, 4 men and 1 group of women can be arranged arround a round table =(5−1)!

Hence, Total arrangements =4!3!=144.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 28

 The total number of sitting arrangements of 7 persons in a row if 2 persons occupy the end seats is _________.

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 29

 A family comprised of an old man, 6 adults and 4 children is to be seated is a row with the condition that the children would occupy both the ends and never occupy either side of the old man. How many sitting arrangements are possible?

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 4 - Question 29

There are 11 persons and 11 seats.

Select 2 children(4C2 ways).
These two children can be seated in the side seats in 2! ways.

The old man can occupy any of the 7 middle seats(7 ways).
(note that he cannot occupy 2nd and 10th seat because these seats are near to the children sitting in the side seats)

3 seats are occupied and 8 seats are remaining. In these 8 seats, 2 seats will be near to the old man in which children cannot seat. Thus 6 seats are remaining in which 2 children can be seated. This can be done in 6P2 ways.

5 persons are seated now.  Remaining 6 persons can be arranged in the remaining 6 seats in 6! ways.

Therefore, required number of ways
= 4C2 × 2! × 7 × 6P2 ×  6!
= 6 × 2 × 7 × 30 × 720
= 1814400

Test: Basic Concepts Of Permutations And Combinations- 4 - Question 30

 In your office 4 posts have fallen vacant. In how many ways a selection out of 31 candidates can be made if one candidate is always included?

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