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Test: Basic Concepts Of Permutations And Combinations- 2


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40 Questions MCQ Test Business Mathematics and Logical Reasoning & Statistics | Test: Basic Concepts Of Permutations And Combinations- 2

Test: Basic Concepts Of Permutations And Combinations- 2 for CA Foundation 2022 is part of Business Mathematics and Logical Reasoning & Statistics preparation. The Test: Basic Concepts Of Permutations And Combinations- 2 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Basic Concepts Of Permutations And Combinations- 2 MCQs are made for CA Foundation 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts Of Permutations And Combinations- 2 below.
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Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 5

The value of 12C4 + 12C3 is

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 6

If npr = 336 and nCr = 56, then n and r will be

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 7

If 18Cr = 18Cr+2, the value of rC5 is

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 8

If ncr–1 = 56, ncr = 28 and ncr+1 = 8, then r is equal to

Test: Basic Concepts Of Permutations And Combinations- 2 - 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.

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 11

 If nc10 = nc14, then 25cn is

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 13

If 28C2r : 24 C2r –4 = 225 : 11, then the value of r is

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 14

The number of diagonals in a decagon is

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 2 - Question 14

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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 15

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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 20

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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 21

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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 22

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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 24

 If 500C92 = 499C92+ nC91 then x is

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 28

The ways of selecting 4 letters from the word EXAMINATION is

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 32

51c31 is equal to

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

*Multiple options can be correct
Test: Basic Concepts Of Permutations And Combinations- 2 - Question 34

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

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 2 - Question 34

COMMERCE : arrangement of 8 words =>

n=8

permutation = n!/(2! * 2!)           [because E and M both repeats 2 times ]

                    = (8*7*6*5*4*3*2!)/(2*1*2!)

                    = 2700

Therefore ,2700 arrangements can be done by the word "COMMERCE".

Test: Basic Concepts Of Permutations And Combinations- 2 - 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.

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 37

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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 38

The number of different factors the number 75600 has is

Detailed Solution for Test: Basic Concepts Of Permutations And Combinations- 2 - Question 38

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

Test: Basic Concepts Of Permutations And Combinations- 2 - Question 39

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

Test: Basic Concepts Of Permutations And Combinations- 2 - 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

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