Test: Basic Concepts Of Permutations And Combinations- 4


40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Basic Concepts Of Permutations And Combinations- 4


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This mock test of Test: Basic Concepts Of Permutations And Combinations- 4 for CA Foundation helps you for every CA Foundation entrance exam. This contains 40 Multiple Choice Questions for CA Foundation Test: Basic Concepts Of Permutations And Combinations- 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Basic Concepts Of Permutations And Combinations- 4 quiz give you a good mix of easy questions and tough questions. CA Foundation students definitely take this Test: Basic Concepts Of Permutations And Combinations- 4 exercise for a better result in the exam. You can find other Test: Basic Concepts Of Permutations And Combinations- 4 extra questions, long questions & short questions for CA Foundation on EduRev as well by searching above.
QUESTION: 1

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

Solution:
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?

Solution:

5C3 x 6C2 = 5 x 4/2 x 6 x 5/2

=150

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?

Solution:

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

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 __________.

Solution:
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?

Solution:
QUESTION: 6

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

Solution:
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.

Solution:
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 ________.

Solution:
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.

Solution:
QUESTION: 10

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

Solution:
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?

Solution:
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 _______.

Solution:
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?

Solution:
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?

Solution:

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

QUESTION: 15

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

Solution:
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?

Solution:

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

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?

Solution:
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_________.

Solution:

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

QUESTION: 19

In how many ways can 8 boys form a ring?

Solution:
QUESTION: 20

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

Solution:
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?

Solution:
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 __________.

Solution:
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.

Solution:
QUESTION: 24

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

Solution:
QUESTION: 25

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

Solution:

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! 

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 __________.

Solution:
QUESTION: 27

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

Solution:

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.

QUESTION: 28

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

Solution:
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?

Solution:

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

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?

Solution:
QUESTION: 31

In a ration shop queue 2 boys, 2 girls, and 2 men are standing in such a way that the boys the girls and the men are together each.The total number of ways of arranging the queue is _______.

Solution:
QUESTION: 32

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

Solution:

For the given case : First fix the women in one group. 3 women can be seated in 3! ways. The 4 men can be seated in the remaining 4 places in 4! ways. Therefore, there are 144 arrangements.

 

Take women as 1 group. So we now have 4 men and 1 group . These 5 men can arranged in round table in (5-1)! = 4! = 24 ways.

But women can be further arranged in group in 3!= 6 ways.

Therefore total number of ways is equal to 24 × 6 = 144 ways .

Case 2- No women sit together.

As number of women is less than number of men , two of the men will be in between a pair of women. So fix the seats of women.

Women can be arranged in 3!= 6 ways.

Now select two men. They can be selected in C(4,2) = 6 ways and arranged in 2! = 2 ways.

Consider these 2 men as a single man. Now we have 3 men and they can be arranged in the round table in (3-1)! = 2! = 2 ways .

Therefore total number of ways = 6 × 6 × 2 × 2 = 144 ways .

QUESTION: 33

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

Solution:
QUESTION: 34

 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?

Solution:
QUESTION: 35

In how many ways can 8 beads of different colour be strung on a ring?

Solution:
QUESTION: 36

 How many numbers greater than a million can be formed with the digits: One 0 Two 1 one 3 and Three 7?

Solution:
QUESTION: 37

In how many ways the letters of the word "LOCKDOWN" can be arranged such that N is always between O's?

Solution:

QUESTION: 38

 How many different words can be formed with the letter of the word “HARIYANA”?

Solution:
QUESTION: 39

 How many number greater than 23000 can be formed with 1, 2, ………5?

Solution:
QUESTION: 40

How many different words can be formed with the letter of the word “HARIYANA”?How many arrangements are possible keeping 'h' and 'n' together ?

Solution: