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## Permutations and Combinations Questions - Permutations and Combinations Quiz Details

In this section, we have discussed Permutations and Combinations Quiz details. The above table states the overview of the Permutations and Combinations MCQ Quiz, the number of questions, exam type, and category. Aspirants who are eagerly waiting to attend the Permutations and Combinations Online Test can check the below portions of this post. Individuals start practicing to get awareness about Permutations and Combinations Aptitude Questions.

### Permutations and Combinations Aptitude Questions

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no time limit allocated for the exam. Moreover, there are no negative marks in the Permutations and Combinations quiz.
### Permutations and Combinations Formulae

1. Factorial Notation - Let n be a positive integer. Then, factorial n denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

2. Number of Permutations - Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!/(n - r)!

3. Then, the number of permutations of these n objects is = n!/(p1!).(p2)!.....(pr!)

4. Number of Combinations - The number of all combinations of n things, taken r at a time is:

nCr = n!/(r!)(n - r)! = n(n - 1)(n - 2) ... to r factors/r!

### Permutations and Combinations MCQ Quiz Answers With Solutions

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Online Test Name |
Permutations and Combinations |

Exam Type |
Multiple Choice Questions |

Category |
Aptitude Quiz |

Number of Questions |
53 |

In this section, we have discussed Permutations and Combinations Quiz details. The above table states the overview of the Permutations and Combinations MCQ Quiz, the number of questions, exam type, and category. Aspirants who are eagerly waiting to attend the Permutations and Combinations Online Test can check the below portions of this post. Individuals start practicing to get awareness about Permutations and Combinations Aptitude Questions.

n! = n(n - 1)(n - 2) ... 3.2.1.

2. Number of Permutations - Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!/(n - r)!

3. Then, the number of permutations of these n objects is = n!/(p1!).(p2)!.....(pr!)

4. Number of Combinations - The number of all combinations of n things, taken r at a time is:

nCr = n!/(r!)(n - r)! = n(n - 1)(n - 2) ... to r factors/r!

31. A question paper had 10 questions. Each question could only be answered as true(T) of False(F). Each candidate answered all the questions, Yet no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?
### Answer & Explanation

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Answer: Option D

Explanation:

Each question can be answered in 2 ways.

10 Questions can be answered = 2^{10}= 1024 ways.

32. Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?
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Answer: Option C

Explanation:

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

Alternate method:

Total number of way in which 3 boys can be selected out of 5 is 5C_{3}

Number of ways in which CD comes together = 3
(CDA,CDB,CDE)

Therefore, Required number of ways = 5C_{3} -3

= 10-3 =7.

33. In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?
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Answer: Option D

Explanation:

1. As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

34. In a question paper, there are four multiple choice type questions, each question has five choices with only one choice for it’s correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?
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Answer: Option C

Explanation:

Multiple choice type questions = 1 2 3 4

Total number of ways = 5x5x5x5 =625.

A number of correct answer = 1.

Number of false answers = 625-1 =624.

35. A mixed doubles tennis game is to be played two teams(each consists of one male and one female) There are four married couples. No team is to consist a husband and his wife. What is the maximum number of games that can be played?
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Answer: Option D

Explanation:

Married couples = MF MF MF MF

AB CD EF GH

Possible teams = AD CB EB GB

AF CF ED GD

AH CH EH GF S

Since one male can be paired with 3 other female, Total teams = 4x3 = 12.

Team AD can play only with CB,CF,CH,EB,EH,GB,GF(7 teams )

Team AD cannot play with AF, AH, ED and GD

The same will apply with all teams, So number of total matches = 12x7 = 84.

But every match includes 2 teams, so the actual number of matches = 84/2 = 42.