1. how many ways can 6 books be arranged on a shelf?
d. 720
(To find the number of ways to arrange the books, you would use the formula for permutations which is n! (n factorial). In this case, since there are 6 books, you would calculate 6! (6 x 5 x 4 x 3 x 2 x 1) which equals 720.)
2. There are 8 girls competing in an ice-skating competition. In how many ways can they finish in first, second, and third place?
a. 336
(For this problem, you would use the permutation formula for picking items from a group: n!/(n-r)! where n is the total number of items and r is the number of items being picked. In this case, n=8 and r=3, so you would calculate 8!/(8-3)! or 8!/(5!) which is equivalent to 8 x 7 x 6 = 336.)
3. How many ways can a president and vice president be selected from a class of 12?
c. 132
(For this problem, n=12 and r=2, so you would calculate 12!/(12-2)! or 12!/(10!) which is equivalent to 12 x 11 = 132.)
4. How many ways can 7 students line up for a class picture?
a. 5,040
(Similar to question 1, you would use the formula for permutations which is n!. In this case, there are 7 students so you would calculate 7! (7 x 6 x 5 x 4 x 3 x 2 x 1) which equals 5,040.)
Connexus Math 7B unit 2 lesson 6 Permutations Quick Check
1. how many ways can 6 books be arranged on a shelf?
a. 21
b. 120
c. 360
d. 720
2. There are 8 girls competing in an ice-skating competition. In how many ways can they finish in first, second, and third place?
a. 336
b. 1,680
c. 2,016
d. 40,320
3. How many ways can a president and vice president be selected from a class of 12?
a. 23
b. 72
c. 132
d. 1,320
4. How many ways can 7 students line up for a class picture?
a. 5,040
b. 210
c. 28
d.7
Please help idk how to find these answers
5 answers
For questions 1-2, simplify each expression.
1. 9!
A.45
B.362,880
C.265,456
D.125,936
2.25P3
A.12,755
B.14,290
C.13,800
D.18.250
3. Which expression would you use to figure out the number of ways you can arrange the letters in the word equation?
A. 8!
B. 8P4
C. 4P8
D.4!
4. Five friends are having their picture taken. How many ways can the photographer arrange the friends in a row?
A.150
B.100
C.120
D.80
5.How would you apply the fundamental counting principle to a bike combination with 5 numbers?
A. 10+10+10+10+10
B. 10 x 5
C. 10 x 10 x 10 x 10 x 10
D. None of these
1. 9!
A.45
B.362,880
C.265,456
D.125,936
2.25P3
A.12,755
B.14,290
C.13,800
D.18.250
3. Which expression would you use to figure out the number of ways you can arrange the letters in the word equation?
A. 8!
B. 8P4
C. 4P8
D.4!
4. Five friends are having their picture taken. How many ways can the photographer arrange the friends in a row?
A.150
B.100
C.120
D.80
5.How would you apply the fundamental counting principle to a bike combination with 5 numbers?
A. 10+10+10+10+10
B. 10 x 5
C. 10 x 10 x 10 x 10 x 10
D. None of these
1. B. 362,880
2. C. 13,800
3. A. 8!
4. C. 120
5. C. 10 x 10 x 10 x 10 x 10
2. C. 13,800
3. A. 8!
4. C. 120
5. C. 10 x 10 x 10 x 10 x 10
Can u tell me how you got the answers?
Yes, sure! Here are the explanations for each question:
1. 9!
To simplify 9!, we just need to evaluate the expression. 9! means 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, which equals 362,880. Therefore, the answer is B. 362,880.
2. 25P3
To simplify 25P3, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, n = 25 and r = 3. So, 25P3 = 25! / (25 - 3)! = 25! / 22! = 25 x 24 x 23 = 13,800. Therefore, the answer is C. 13,800.
3. Which expression would you use to figure out the number of ways you can arrange the letters in the word equation?
To find the number of ways to arrange the letters in the word equation, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, there are 8 letters in the word equation, so we need to find the number of permutations of 8 things taken 8 at a time. Therefore, the answer is A. 8!.
4. Five friends are having their picture taken. How many ways can the photographer arrange the friends in a row?
To find the number of ways to arrange the friends in a row, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, there are 5 friends, so we need to find the number of permutations of 5 things taken 5 at a time. Therefore, the answer is C. 120.
5. How would you apply the fundamental counting principle to a bike combination with 5 numbers?
The fundamental counting principle states that the total number of outcomes of a sequence of events is equal to the product of the number of ways each event can occur. In this case, we have 5 numbers on a bike combination and each number can be any digit from 0 to 9. Therefore, the number of possible combinations is 10 x 10 x 10 x 10 x 10, or 10^5. Therefore, the answer is C. 10 x 10 x 10 x 10 x 10.
1. 9!
To simplify 9!, we just need to evaluate the expression. 9! means 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, which equals 362,880. Therefore, the answer is B. 362,880.
2. 25P3
To simplify 25P3, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, n = 25 and r = 3. So, 25P3 = 25! / (25 - 3)! = 25! / 22! = 25 x 24 x 23 = 13,800. Therefore, the answer is C. 13,800.
3. Which expression would you use to figure out the number of ways you can arrange the letters in the word equation?
To find the number of ways to arrange the letters in the word equation, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, there are 8 letters in the word equation, so we need to find the number of permutations of 8 things taken 8 at a time. Therefore, the answer is A. 8!.
4. Five friends are having their picture taken. How many ways can the photographer arrange the friends in a row?
To find the number of ways to arrange the friends in a row, we need to use the permutation formula: nPr = n! / (n - r)!. In this case, there are 5 friends, so we need to find the number of permutations of 5 things taken 5 at a time. Therefore, the answer is C. 120.
5. How would you apply the fundamental counting principle to a bike combination with 5 numbers?
The fundamental counting principle states that the total number of outcomes of a sequence of events is equal to the product of the number of ways each event can occur. In this case, we have 5 numbers on a bike combination and each number can be any digit from 0 to 9. Therefore, the number of possible combinations is 10 x 10 x 10 x 10 x 10, or 10^5. Therefore, the answer is C. 10 x 10 x 10 x 10 x 10.