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Test: Permutations and Combinations- 2 - CA Foundation MCQ


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30 Questions MCQ Test - Test: Permutations and Combinations- 2

Test: Permutations and Combinations- 2 for CA Foundation 2024 is part of CA Foundation preparation. The Test: Permutations and Combinations- 2 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Permutations and Combinations- 2 MCQs are made for CA Foundation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutations and Combinations- 2 below.
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Test: 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: 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: Permutations and Combinations- 2 - Question 2

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Test: 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: 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: Permutations and Combinations- 2 - Question 5

The value of 12C4 + 12C3 is

Test: Permutations and Combinations- 2 - Question 6

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

Detailed Solution for Test: Permutations and Combinations- 2 - Question 6

if ncr=56 and npr=336 find n

Test: Permutations and Combinations- 2 - Question 7

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

Test: Permutations and Combinations- 2 - Question 8

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

Test: 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.

Detailed Solution for Test: Permutations and Combinations- 2 - Question 9

For each friend, the person has two choices: either to invite them or not to invite them. Since there are 8 friends, each with 2 choices, the total number of ways to choose friends (including choosing none) is:

28 = 256
However, since the question specifies "one or more" friends, we subtract the case where no friends are invited.
256-1 = 255

Correct answer: b) 255

Test: 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: Permutations and Combinations- 2 - Question 11

 If nc10 = nc14, then 25cn is

Test: 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: Permutations and Combinations- 2 - Question 13

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

Test: Permutations and Combinations- 2 - Question 14

The number of diagonals in a decagon is

Detailed Solution for Test: Permutations and Combinations- 2 - Question 14

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

Test: Permutations and Combinations- 2 - Question 15

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

Detailed Solution for Test: Permutations and Combinations- 2 - Question 15

Explanation

- Given that there are 12 points in a plane.
- Out of these 12 points, 5 are collinear.
- To find the number of triangles that can be formed using these 12 points, we can use the combination formula.
- The number of ways to choose 3 points out of 12 is given by 12C3 = 220.
- However, we need to subtract the number of triangles that can be formed using the collinear points.
- Since 5 points are collinear, we can form triangles using these points as well.
- The number of ways to choose 3 points out of 5 collinear points is given by 5C3 = 10.
- So, the total number of triangles that can be formed using the 12 points is 220 - 10 = 210.
- Therefore, the correct answer is option C: 210.

Test: 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

Detailed Solution for Test: Permutations and Combinations- 2 - Question 16

To find the number of straight lines that can be formed by 16 non-collinear points in a plane, we can use the concept of combinations. Here’s the step-by-step solution:

Step 1: Understanding the Problem
We need to find the number of straight lines that can be formed by joining any two points from a set of 16 non-collinear points. Since no three points are collinear, any two points will form a unique straight line.

Step 2: Using Combinations
To determine how many ways we can choose 2 points from the 16 points, we use the combination formula:

In this case, n=16 (the total number of points) and r=2 (the number of points we are choosing to form a line).

Step 3: Applying the Combination Formula
Substituting the values into the formula gives us:

Step 5: Calculating 2!
Calculating 2!:

2!=2×1=2

Step 6: Final Calculation
Now substituting back into our equation:

Test: 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

Detailed Solution for Test: Permutations and Combinations- 2 - Question 17


Test: 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: 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: Permutations and Combinations- 2 - Question 20

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

Test: Permutations and Combinations- 2 - Question 21

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

Test: 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: 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: Permutations and Combinations- 2 - Question 24

 If 500C92 = 499C92+ nC91 then x is

Detailed Solution for Test: Permutations and Combinations- 2 - Question 24

Test: 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: 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: 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: Permutations and Combinations- 2 - Question 28

The ways of selecting 4 letters from the word EXAMINATION is

Test: 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: 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

Detailed Solution for Test: Permutations and Combinations- 2 - Question 30

Let the two particular guests sit on right side.
So the three particular guests will sit on left side.

So remaining will be 3 people which need to be selected.
From these 3 people 2 will sit on right side and the one will sit on left side.

Total ways of selecting the remaining will be = ³C₂ ⨯ ¹C₁ = 3

Total ways of arranging the people will be =

Selection of remaining × 4! ( for arranging people on left side) × 4! ( arranging people on right side)
= 3 × 24 × 24 = 3 × 576 = 1728

So in 1728 ways we can arrange them

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