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# Test: Basic Concepts Of Permutations And Combinations- 3

## 40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Basic Concepts Of Permutations And Combinations- 3

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This mock test of Test: Basic Concepts Of Permutations And Combinations- 3 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- 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Basic Concepts Of Permutations And Combinations- 3 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- 3 exercise for a better result in the exam. You can find other Test: Basic Concepts Of Permutations And Combinations- 3 extra questions, long questions & short questions for CA Foundation on EduRev as well by searching above.
QUESTION: 1

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QUESTION: 2

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QUESTION: 3

### If six times the number of permutations of n items taken 3 at a time is equal to seven times the number of permutation of (n-1) items taken 3 at a time, then the value of n will be:

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QUESTION: 4

The number of words that can be formed out of the letters of the word "ARTICLE" so that vowels occupy even places is:

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QUESTION: 5

If 1000C98 = 999C97 + xc901 , then the value of x will be:

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QUESTION: 6

Number of ways of shaking hands in a group of 10 persons shaking hands to each other are:

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QUESTION: 7

If 15C3r = 15Cr+3, then r is equal is

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QUESTION: 8

If six times the number of permutations of n items taken 3 at a time is equal to seven times the number of permutation of (n-1) items taken 3 at a time, then the value of n will be:

Solution:
QUESTION: 9

The ways of selecting 4 letters from the word EXAMINATION is

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QUESTION: 10

The letters of the word "VIOLENT" are arranged so that the vowels occupy even place only. The number of permutations is ______.

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QUESTION: 11

How many permutations can be formed from the letters of the word  " DRAUGHT" if both vowels may not be separated?

Solution:

The vowels are never being separated.

So, vowels a & u are taken together and considered as one letter.

So, there are 6 letters in the word draught

The number of ways in which it can be arranged are given by =6!=720.

But the internal arrangement can be done of a and u in 2 ways.

Hence the total arrangements are 720×2=1440.

QUESTION: 12

If nP4 = 20 (nP2) then the value of n is ______________.

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QUESTION: 13

A polygon has 44 diagonals then the number of its sides are:

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QUESTION: 14

How many different words can be formed with the letters of the word "LIBERTY"

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QUESTION: 15

You have to make a choice of 4 balls out of one red one blue and ten white balls. The numbers of  ways in which  this can be done to include the red ball  but exclude the blue ball always is ________.

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QUESTION: 16

You have to make a choice of 4 balls out of one red one blue and ten white balls. The numbers of ways this can be done to always include the red ball is _______.

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QUESTION: 17

In a cross word puzzle 20 words are to be guessed of which 8 words have each an alternative solution. The number of possible solution is ________.

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QUESTION: 18

How many combinations can be formed of eight counters marked 1, 2, 3, 4, 5, 6, 7, 8 taking them 4 at a time, there being at least one odd and one even counter in each combination?

Solution:

We have,1,2,3,4,5,6,7,8

Total counters = 8

Taken at a time = 4

The total combinations without restrictions= 8C4 = 70

Odd = 1,3,5,7

Even = 2,4,6,8

We have to find out those combinations which contain at least one even and one odd counter i.e. those neither can have all 4 even counters nor all 4 odd counters.

Let's first find the the no. of combinations with restrictions i.e. which contains all even or all odd counters.

So, the no. of combination which have all 4 odd counters = 4C4 = 1

Similarly, the no. of combination which have all 4 even counters = 4C4 = 1

Hence, the required no. combinations which contain at least one even and one odd counters = Total - Restricted combinations

= 70–1–1

= 68 Ans.

QUESTION: 19

Find the number of ways in which a selection of four letters can be made from the letters of the word 'PROPORTION,

Solution:

Given the word is

PROPORTION

When

Number of P is =2

Number of O is =3

Number of R is =2

Number of I is =1

Number of T is =1

Number of N is =1

Let us frist find the number of selections.

Case 1:- All four letters distinct = number of ways will be 4×6C4​=4!×4!(6−4)!6!​=2!6×5×4×3×2×1​=360

Case 2:- Three letters same and one letter distinct, then no. of ways  = 3!4!​5C1​=3!4×3!​×5=20

Case 3:- Two letters of one type and the other two letters of other type, then no. of ways(out for P,R and O)=3C2​×4C2​=3×2!×2!4!​=18

Case 4:-two letters same and other two letters are different.then, no. of ways. = 3C1​×5C2​=3×10=30×2!4!​=360

Now according to given question,

Total number of selections=15+5+3+30=53

QUESTION: 20

Out of 6 members belonging to party

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QUESTION: 21

A team of 5 is to be selected from 8 boys and three girls. Find the probability that it includes two particular girls.

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QUESTION: 22

You have to make a choice of 4 balls out of one red one blue and ten white balls. The number of ways in which this can be done to exclude both the red and blues ball is ________.

Solution:

Total no. of balls =12

They said to exclude both red&blue
then, remaining total balls=10
out of this 4 should be selected

QUESTION: 23

The number of words which can be formed with 2 different consonants and 1 vowel out of 7 different consonants and 3 different vowels the vowel to lie between 2 consonants is _________.

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QUESTION: 24

A question paper divided into 2 groups consisting of 3 and 4 questions respectively carries the note

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QUESTION: 25

Find the number of ways in which an arrangement of 4 letters can be made from the word

Solution:

The word contain 11 letters with repetition of letters.
We can choose 4 letters by
i) all the four letters are distinct
ii) Two distinct and two alike
iii) Two alike of one kind and two alike of other kind
i) 8 different letters are M, A, T, H, E, I, C, S
This can be chosen by 8C4 × 4! = 1680
ii) Three pairs alike letters MM, AA, TT. One pair can be chosen by 3C1 = 3 ways
Remaining two letters can be chosen by 7C
Each such groups has 4 letters out of which 2 are alike.
Hence, it can be chosen by 4!/2!
Total number of ways = 3 × 7C2 × 4!/2! = 756
iii) Three pairs of 2 alike letters. Two pairs can be chosen by 3C2
3C2 groups of 4 letters each.
There are 4 letters of which 2 alike of one kind and two alike of other kind in each group.
That can be arranged by 4!/2! × 1/2!
Total number of ways = 3C2 × 4!/2! × 1/2! = 18
Total number of ways = 1680 + 756 + 18 = 2454

QUESTION: 26

Find  the number of arrangements of 5 things taken out of 12 things, in which one particular thing must always be included.

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QUESTION: 27

In how many ways 3 prizes out of 5 can be distributed amongst 3 brothers Equally?

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QUESTION: 28

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?

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QUESTION: 29

In how many ways 21 red balls and 19 blue balls can be arranged in a row so that no two blue balls are together?

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QUESTION: 30

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? How many choices you have to make if there are 2 males?

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QUESTION: 31

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? How many choices you have to make if there is at least one female?

Solution:

5 males and 6 females = 11 people. A committee of 5 out of 11 people can be selected in 11C5 =462 ways. At least female basically means that it can’t be an all-committee group. In this case, there is only 1 way to make an all-male committee, so just remove that case and you have 462–1=461 ways to choose a committee with at least one female.

QUESTION: 32

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? How many choices you have to make if there are 2 males?

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QUESTION: 33

A committee is to be formed of 2 teachers and 3 students out of 10 teachers and 20 students. If a particular student is excluded the number of ways in which this can be done is _________.

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QUESTION: 34

In how many ways you can answer one or more questions out of 6 questions each having an alternative?

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QUESTION: 35

From 7 men and 4 women a committee of 5 is to be formed. In how many ways can this be done to include at least one woman?

Solution:
QUESTION: 36

A committee is to be formed of 2 teachers and 3 students out of 10 teachers and 20 students. The number of ways in which this can be done is ___________.

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QUESTION: 37

A committee is to be formed of 2 teachers and 3 students out of 10 teachers and 20 students. If a particular teacher is included the number of ways in which this can be done is ________.

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QUESTION: 38

There are 12 points in a plane no 3 of which are collinear except that 6 points which are collinear. The number of different straight lines is ________.

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QUESTION: 39

There are 12 points in a plane no 3 of which are collinear except that 6 points which are collinear. The number of different triangles formed by joining the straight lines is __________.

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QUESTION: 40

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? How many choices you have to make if there is no female?

Solution: