Description

This mock test of Test: Basic Concepts Of Permutations And Combinations- 2 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- 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Basic Concepts Of Permutations And Combinations- 2 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- 2 exercise for a better result in the exam. You can find other Test: Basic Concepts Of Permutations And Combinations- 2 extra questions,
long questions & short questions for CA Foundation on EduRev as well by searching above.

QUESTION: 1

The number of arrangements in which the letters of the word MONDAY be arranged so that the words thus formed begin with M and do not end with N is

Solution:

QUESTION: 2

The total number of ways in which six ‘+’ and four ‘–‘ signs can be arranged in a line such that no two ‘–’ signs occur together is

Solution:

QUESTION: 3

The number of ways in which the letters of the word MOBILE be arranged so that consonants always occupy the odd places is

Solution:

QUESTION: 4

5 persons are sitting in a round table in such way that Tallest Person is always on the right– side of the shortest person; the number of such arrangements is

Solution:

QUESTION: 5

The value of ^{12}C_{4} + ^{12}C_{3} is

Solution:

QUESTION: 6

If ^{n}p_{r} = 336 and ^{n}C_{r} = 56, then n and r will be

Solution:

QUESTION: 7

If ^{18}C_{r} = ^{18}C_{r+2}, the value of^{ r}C_{5} is

Solution:

QUESTION: 8

If^{ n}c_{r–1} = 56, ^{n}c_{r} = 28 and ^{n}c_{r+1} = 8, then r is equal to

Solution:

QUESTION: 9

A person has 8 friends. The number of ways in which he may invite one or more of them to a dinner is.

Solution:

QUESTION: 10

The number of ways in which a person can chose one or more of the four electrical appliances : T.V, Refrigerator, Washing Machine and a cooler is

Solution:

QUESTION: 11

If ^{n}c_{10} = ^{n}c_{14}, then ^{25}c_{n} is

Solution:

QUESTION: 12

Out of 7 gents and 4 ladies a committee of 5 is to be formed. The number of committees such that each committee includes at least one lady is

Solution:

QUESTION: 13

If ^{28}C_{2r }: ^{24} C_{2r –4} = 225 : 11, then the value of r is

Solution:

QUESTION: 14

The number of diagonals in a decagon is

Solution:

Hint: The number of diagonals in a polygon of n sides is 1/2n (n–3).

QUESTION: 15

There are 12 points in a plane of which 5 are collinear. The number of triangles is

Solution:

QUESTION: 16

The number of straight lines obtained by joining 16 points on a plane, no twice of them being on the same line is

Solution:

QUESTION: 17

At an election there are 5 candidates and 3 members are to be elected. A voter is entitled to vote for any number of candidates not greater than the number to be elected. The number of ways a voter choose to vote is

Solution:

QUESTION: 18

Every two persons shakes hands with each other in a party and the total number of hand shakes is 66. The number of guests in the party is

Solution:

QUESTION: 19

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is

Solution:

QUESTION: 20

The number of ways in which 12 students can be equally divided into three groups is

Solution:

QUESTION: 21

The number of ways in which 15 mangoes can be equally divided among 3 students is

Solution:

QUESTION: 22

8 points are marked on the circumference of a circle. The number of chords obtained by joining these in pairs is

Solution:

QUESTION: 23

A committee of 3 ladies and 4 gents is to be formed out of 8 ladies and 7 gents. Mrs. X refuses to serve in a committee in which Mr. Y is a member. The number of such committees is

Solution:

QUESTION: 24

If ^{500}C_{92} = ^{499}C_{92}+ ^{n}C_{91} then x is

Solution:

QUESTION: 25

The Supreme Court has given a 6 to 3 decision upholding a lower court; the number of ways it can give a majority decision reversing the lower court is

Solution:

QUESTION: 26

Five bulbs of which three are defective are to be tried in two bulb points in a dark room.Number of trials the room shall be lighted is

Solution:

QUESTION: 27

The letters of the words CALCUTTA and AMERICA are arranged in all possible ways.The ratio of the number of there arrangements is

Solution:

QUESTION: 28

The ways of selecting 4 letters from the word EXAMINATION is

Solution:

QUESTION: 29

The number of different words that can be formed with 12 consonants and 5 vowels by taking 4 consonants and 3 vowels in each word is

Solution:

QUESTION: 30

Eight guests have to be seated 4 on each side of a long rectangular table.2 particular guests desire to sit on one side of the table and 3 on the other side. The number of ways in which the sitting arrangements can be made is

Solution:

QUESTION: 31

A question paper contains 6 questions, each having an alternative.

The number of ways an examine can answer one or more questions is

Solution:

QUESTION: 32

^{51}c_{31} is equal to

Solution:

QUESTION: 33

The number of words that can be made by rearranging the letters of the word APURNA so that vowels and consonants appear alternate is

Solution:

*Multiple options can be correct

QUESTION: 34

The number of arrangement of the letters of the word COMMERCE is

Solution:

QUESTION: 35

A candidate is required to answer 6 out of 12 questions which are divided into two groups containing 6 questions in each group. He is not permitted to attempt not more than four from any group. The number of choices are.

Solution:

QUESTION: 36

The results of 8 matches (Win, Loss or Draw) are to be predicted. The number of different forecasts containing exactly 6 correct results is

Solution:

QUESTION: 37

The number of ways in which 8 different beads be strung on a necklace is

Solution:

QUESTION: 38

The number of different factors the number 75600 has is

Solution:

75600 = 756*100 = 7*108*100 =7*4*27*100=7*(3^3)*(4)*(4)*(5)*(5) = 7*(3^3)*(2^4)*(5^2)

thus the powers of the prime factors are, 1, 3, 4, 2

thus number of factors are,

(1+1)(1+3)(1+4)(1+2) = 120

QUESTION: 39

The number of 4 digit numbers formed with the digits 1, 1, 2, 2, 3, 4 is

Solution:

QUESTION: 40

The number of ways a person can contribute to a fund out of 1 ten-rupee note, 1 fiverupee note, 1 two-rupee and 1 one rupee note is

Solution:

### Basic Concepts of Permutations and Combinations

Doc | 38 Pages

### Permutations & Combinations

Doc | 34 Pages

### Permutation theorems - Permutations & Combinations

Video | 13:30 min

### PPT - Permutations & Combinations

Doc | 49 Pages

- Test: Basic Concepts Of Permutations And Combinations- 2
Test | 40 questions | 40 min

- Test: Basic Concepts Of Permutations And Combinations- 4
Test | 40 questions | 40 min

- Test: Basic Concepts Of Permutations And Combinations- 1
Test | 40 questions | 40 min

- Test:- Permutations And Combinations - 2
Test | 20 questions | 60 min

- Permutations And Combinations - MCQ 2
Test | 20 questions | 15 min