Counting Principle And Permutation Quiz

1. How many ways can 10 students win a race?

2

5

10

13

There are 10 students participating in the race, and each student has a chance of winning. Therefore, there are 10 possible ways in which the students can win the race.

Explanation

There are 10 students participating in the race, and each student has a chance of winning. Therefore, there are 10 possible ways in which the students can win the race.

2. How many ways can we arrange 7 people sitting on 3 chairs?

112

157

173

210

There are 7 people and 3 chairs available. The first person has 7 options to choose from, the second person has 6 options remaining, and the third person has 5 options left. Therefore, the total number of ways to arrange the 7 people on 3 chairs is 7 x 6 x 5 = 210.

Explanation

There are 7 people and 3 chairs available. The first person has 7 options to choose from, the second person has 6 options remaining, and the third person has 5 options left. Therefore, the total number of ways to arrange the 7 people on 3 chairs is 7 x 6 x 5 = 210.

3. How many ways can I arrange the letters of the word FLORIDA?

5010

5020

5040

5060

To find the number of ways to arrange the letters in the word "FLORIDA," we consider the total number of arrangements, which is 7 factorial (7!). This means multiplying 7 by each decreasing positive integer down to 1.

So, it's like calculating 7 times 6 times 5 times 4 times 3 times 2 times 1, which equals 5040. Therefore, there are 5040 different ways you can arrange the letters in the word "FLORIDA."

Explanation

To find the number of ways to arrange the letters in the word "FLORIDA," we consider the total number of arrangements, which is 7 factorial (7!). This means multiplying 7 by each decreasing positive integer down to 1.

So, it's like calculating 7 times 6 times 5 times 4 times 3 times 2 times 1, which equals 5040. Therefore, there are 5040 different ways you can arrange the letters in the word "FLORIDA."

4. I can choose a red, blue, or green T-shirt, with a skirt or pants, a black, white, or green hijab, with or without a jacket. How many different choices do I have?

3 x 2 x 3 x 2

3 x 1 x 2 x 3

3 x 3x 0 x 1

4 x 9 x 5x 3

There are 3 choices for the color of the T-shirt (red, blue, or green), 2 choices for the type of bottom (skirt or pants), 3 choices for the color of the hijab (black, white, or green), and 2 choices for whether or not to wear a jacket. Therefore, the total number of different choices is calculated by multiplying these numbers together: 3 x 2 x 3 x 2 = 36.

Explanation

There are 3 choices for the color of the T-shirt (red, blue, or green), 2 choices for the type of bottom (skirt or pants), 3 choices for the color of the hijab (black, white, or green), and 2 choices for whether or not to wear a jacket. Therefore, the total number of different choices is calculated by multiplying these numbers together: 3 x 2 x 3 x 2 = 36.

5. From 15 entries in an art contest, a camp counselor chooses first, second, and third place winners. How many choices can be made?

2730

15

45

8210

There are 15 choices for the first place winner, then 14 choices remaining for the second place winner, and 13 choices remaining for the third place winner. Therefore, the total number of choices that can be made is 15 * 14 * 13 = 2730.

Explanation

There are 15 choices for the first place winner, then 14 choices remaining for the second place winner, and 13 choices remaining for the third place winner. Therefore, the total number of choices that can be made is 15 * 14 * 13 = 2730.

6. How many ways can I arrange 5 books?

5 ways

10 ways

120 ways

150 ways

The number of ways to arrange 5 books can be calculated using the concept of permutations. Since the order of arrangement matters, we can use the formula for permutations of n objects taken r at a time, which is n! / (n-r)!. In this case, we have 5 books to arrange, so the formula becomes 5! / (5-5)! = 5! / 0! = 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Therefore, the correct answer is 120 ways.

Explanation

The number of ways to arrange 5 books can be calculated using the concept of permutations. Since the order of arrangement matters, we can use the formula for permutations of n objects taken r at a time, which is n! / (n-r)!. In this case, we have 5 books to arrange, so the formula becomes 5! / (5-5)! = 5! / 0! = 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Therefore, the correct answer is 120 ways.

7. Evaluate P(5,3)

20

60

57

82

The question is asking to evaluate the permutation of 5 objects taken 3 at a time. The formula for evaluating permutations is P(n,r) = n! / (n-r)!. In this case, P(5,3) = 5! / (5-3)! = 5! / 2! = 5 x 4 x 3 / 2 x 1 = 60. Therefore, the correct answer is 60.

Explanation

The question is asking to evaluate the permutation of 5 objects taken 3 at a time. The formula for evaluating permutations is P(n,r) = n! / (n-r)!. In this case, P(5,3) = 5! / (5-3)! = 5! / 2! = 5 x 4 x 3 / 2 x 1 = 60. Therefore, the correct answer is 60.

8. How many 6-character passwords can be formed if the first character is a digit and the remaining 5 characters are letters that can be repeated?

10x26x25x24x23x22

10x26x26x26x26x26

10x26!

260

The question asks for the number of 6-character passwords that can be formed. The first character must be a digit, which gives us 10 choices (0-9). The remaining 5 characters can be any letter, which gives us 26 choices for each character. Since the letters can be repeated, we multiply the choices together: 10x26x26x26x26x26. This gives us the total number of possible passwords that can be formed.

Explanation

The question asks for the number of 6-character passwords that can be formed. The first character must be a digit, which gives us 10 choices (0-9). The remaining 5 characters can be any letter, which gives us 26 choices for each character. Since the letters can be repeated, we multiply the choices together: 10x26x26x26x26x26. This gives us the total number of possible passwords that can be formed.

9. How many different license plates are possible if the first two places are numbers and the last two are letters, with repeating numbers or letters?

92942

29823

57291

67600

The given answer, 67600, is the number of different license plates that are possible if the first two places are numbers and the last two are letters, with repeating numbers or letters. This can be calculated by multiplying the number of possibilities for each place. For the first two places as numbers, there are 10 options (0-9). For the last two places as letters, there are 26 options (A-Z). Since repetition is allowed, the total number of possibilities is 10 * 10 * 26 * 26 = 67600.

Explanation

The given answer, 67600, is the number of different license plates that are possible if the first two places are numbers and the last two are letters, with repeating numbers or letters. This can be calculated by multiplying the number of possibilities for each place. For the first two places as numbers, there are 10 options (0-9). For the last two places as letters, there are 26 options (A-Z). Since repetition is allowed, the total number of possibilities is 10 * 10 * 26 * 26 = 67600.

10. A briefcase lock has 3 rotating cylinders, each containing 10 digits. How many numerical codes are possible?

10

30

1000

1200

Each cylinder can be set to any of the 10 digits, so there are 10 options for each cylinder. Since there are 3 cylinders, the total number of possible combinations is 10 x 10 x 10 = 1000.

Explanation

Each cylinder can be set to any of the 10 digits, so there are 10 options for each cylinder. Since there are 3 cylinders, the total number of possible combinations is 10 x 10 x 10 = 1000.