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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!

26. A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group?
### Answer & Explanation

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Answer: Option B

Explanation:

Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways

= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5

= 600 ways.

27. A group consists of 4 men, 6 women and 5 children. In how many ways can 3 men, 2 women and 3 children selected from the given group?
### Answer & Explanation

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Answer: Option C

Explanation:

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃

= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)

= 4 * 15 * 10

= 600 ways.

28. A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?
### Answer & Explanation

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Answer: Option C

Explanation:

Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅

= ¹²C₅ = ¹²C₇

29. A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior?
### Answer & Explanation

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Answer: Option D

Explanation:

The total number of ways of forming the group of ten representatives is ²²C₁₀.

The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way

The required number of ways = ²²C₁₀ - 1

30. Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
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Answer: Option E

Explanation:

We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.

Hence, the required number of straight lines

= ¹¹C₂ - ⁶C₂ - ⁵C₂ + 1 + 1

= 55 - 15 - 10 + 2 = 32