# Permutations And Combinations MCQ Quiz Questions With Answers

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What do you understand by permutation and combination? Can you attempt these Permutations and Combinations MCQ Quiz Questions With Answers? If so, take this quiz and check how good you are at solving such word problems. A permutation is an act of arranging the objects in order. A combination is a way of selecting the objects from a group irrespective of the order of the objects. Here, in the quiz below, we will test your logical skills. Try the quiz, then. Good luck!

• 1.

### In how many ways can 3 people be seated in a row containing 6 seats?

• A.

110

• B.

120

• C.

130

• D.

140

B. 120
Explanation
There are 6 seats available and 3 people to be seated. The first person can choose any of the 6 seats. After the first person is seated, there are 5 seats remaining. The second person can choose any of the 5 seats. After the second person is seated, there are 4 seats remaining. The third person can choose any of the 4 seats. Therefore, the total number of ways to seat 3 people in a row containing 6 seats is 6 x 5 x 4 = 120.

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• 2.

### How many 4-letter codes can be formed using the first 9 letters of the English alphabet if no letter is repeated?

• A.

3024

• B.

3036

• C.

3021

• D.

3034

A. 3024
Explanation
To form a 4-letter code using the first 9 letters of the English alphabet without repetition, we have 9 choices for the first letter, 8 choices for the second letter, 7 choices for the third letter, and 6 choices for the fourth letter. Therefore, the total number of possible codes is 9 x 8 x 7 x 6 = 3024.

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• 3.

### Find the number of ways in which 5 players out of 8 players can be selected from a team.

• A.

56

• B.

112

• C.

64

• D.

128

A. 56
Explanation
The number of ways to select 5 players out of 8 can be calculated using the combination formula. The formula for combination is nCr = n! / (r!(n-r)!), where n is the total number of players and r is the number of players to be selected. In this case, n = 8 and r = 5. Plugging these values into the formula, we get 8C5 = 8! / (5!(8-5)!) = 8! / (5!3!) = (8*7*6) / (3*2*1) = 56. Therefore, the correct answer is 56.

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• 4.

### A student has 5 pants and 8 shirts. The number of ways in which he can wear the dress in different combinations is:

• A.

200

• B.

13

• C.

40

• D.

None of the above

C. 40
Explanation
To find the number of ways the student can wear the dress in different combinations, you multiply the number of options for pants by the number of options for shirts.
Number of ways = (number of pants) * (number of shirts) Number of ways = 5 * 8 Number of ways = 40
So, the student can wear the dress in 40 different combinations.

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• 5.

### There are 35 teachers in a school. In how many different ways one principal and one vice principal can be chosen?

• A.

1160

• B.

1170

• C.

1180

• D.

1190

D. 1190
Explanation
The number of ways to choose one principal from 35 teachers is 35. After selecting the principal, there are 34 teachers remaining from which to choose the vice principal. Therefore, the number of ways to choose one vice principal is 34. To find the total number of ways to choose both a principal and a vice principal, we multiply the number of ways to choose each position together: 35 * 34 = 1190.

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• 6.

### Four boys and three girls are to be seated for a dinner such that no two girls sit together and no two boys sit together. Find the number of ways in which this can be arranged.

• A.

36

• B.

72

• C.

144

• D.

180

C. 144
Explanation
In this scenario, we need to arrange four boys and three girls for a dinner in such a way that no two girls sit together and no two boys sit together. To find the number of ways this can be arranged, we can consider the arrangement of boys and girls separately. There are 4! (4 factorial) ways to arrange the boys and 3! (3 factorial) ways to arrange the girls. However, since the arrangement of boys and girls is independent, we need to multiply these two numbers together. Therefore, the total number of ways to arrange the boys and girls is 4! * 3! = 24 * 6 = 144.

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• 7.

• A.

24

• B.

576

• C.

144

• D.

440

C. 144
• 8.

• A.

209

• B.

159

• C.

194

• D.

205

A. 209
• 9.

### How many 4-letter words, with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’ if repetition of letters is not allowed?

• A.

40

• B.

400

• C.

5040

• D.

2520

C. 5040
Explanation
The given word "LOGARITHMS" has 10 letters. We need to form 4-letter words without repetition. To do this, we can select 4 letters from the 10 available. The number of ways to select 4 letters from 10 is given by the combination formula, which is 10! / (4! * (10-4)!). Simplifying this expression gives us 10! / (4! * 6!). This can be further simplified to (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 5040. Therefore, there are 5040 possible 4-letter words that can be formed.

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• 10.

### How many ways can the letters of the word TREES be ordered such that each ‘word’ starts with a consonant and ends with a vowel?

• A.

9

• B.

24

• C.

27

• D.

18

D. 18
Explanation
To determine the number of ways the letters of the word TREES can be arranged such that each "word" starts with a consonant and ends with a vowel, we need to consider the possible arrangements. The word TREES has 5 letters, with 2 vowels (E, E) and 3 consonants (T, R, S).

To start with a consonant, we have 3 options for the first letter. Then, for the second position, we have 4 options (as we have used one of the consonants already). For the third position, we have 3 options left, and for the fourth position, we have 2 options. Finally, for the last position, we have 2 options for a vowel.

Hence, the total number of ways the letters can be arranged is 3 x 4 x 3 x 2 x 2 = 72. However, we need to divide by 2 to account for the repeated vowel (E), resulting in 72/2 = 36.

Therefore, the correct answer is 18.

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• 11.

### 12 people at a party shake hands once with everyone else in the room. How many handshakes took place?

• A.

132

• B.

66

• C.

12! / 2

• D.

12!

B. 66
Explanation
In this scenario, each person shakes hands with every other person in the room once. To calculate the number of handshakes, we can use the formula n(n-1)/2, where n represents the number of people. In this case, there are 12 people, so the calculation would be 12(12-1)/2 = 66 handshakes.

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• 12.

### In how many ways can you choose one or more of 5 different candies?

• A.

16

• B.

21

• C.

31

• D.

32

C. 31
Explanation
The number of ways to choose one or more of 5 different candies can be found using the concept of combinations. For each candy, we have two choices: either include it in the selection or not. Since we have 5 candies, the total number of choices is 2^5 = 32. However, we need to subtract 1 from this total because we are not considering the case where none of the candies are chosen. Therefore, the correct answer is 32 - 1 = 31.

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• 13.

### In how many ways can 8 directors, vice-chairman, and chairman of a firm be selected at a round table if the chairman has to sit between the vice-chairman and a director?

• A.

9! * 2

• B.

8! * 2

• C.

2! * 7

• D.

8!

• E.

9!

B. 8! * 2
Explanation
There are 8 ways to select the director who will sit on the left side of the chairman, and once that is determined, there is only 1 way to select the chairman. Then, there are 7 ways to select the director who will sit on the right side of the chairman. Therefore, the total number of ways to select the directors, vice-chairman, and chairman is 8 * 1 * 7 = 56. However, since the directors and vice-chairman can be arranged in a circular manner, we need to multiply this by 2. Thus, the correct answer is 8! * 2.

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• 14.

### A man has 9 friends, 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitees?

• A.

320

• B.

160

• C.

80

• D.

200

B. 160
Explanation
The man has 5 girls to choose from for the first girl to be invited, then 4 girls for the second girl, and 3 girls for the third girl. Since the order in which the girls are invited does not matter, we divide by 3! (the number of ways to arrange 3 girls) to avoid overcounting. For the remaining 6 spots, the man can invite any of the 4 boys or the remaining 2 girls. Therefore, the total number of ways to invite the friends is 5 * 4 * 3 / 3! * 6 * 5 = 160.

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• 15.

### In how many ways can you distribute 5 identical chocolates to 3 children so that any child can get any number of chocolates from 0 to 5?

• A.

21

• B.

42

• C.

5! / 3!

• D.

5! / 3!

• E.

60

A. 21
Janaisa Harris |BA (Mathematics) |
High School Math Teacher
Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a bachelor's degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher.

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• Jul 05, 2017
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