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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!

1. A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
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Answer: Option D

Explanation:

The boy can select one trouser in nine ways.

The boy can select one shirt in 12 ways.

The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

2. How many three letter words are formed using the letters of the word TIME?
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Answer: Option D

Explanation:

The number of letters in the given word is four.

The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

3. Using all the letters of the word "THURSDAY", how many different words can be formed?
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Answer: Option B

Explanation:

Total number of letters = 8

Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.

4. Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
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Answer: Option B

Explanation:

There are five letters in the given word.

Consider 5 blanks ....

The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.

The number of words = 3! = 6.

5. The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?
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Answer: Option B

Explanation:

The word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144.