Question 1

Solve Counting Problems Using Combinations
-A professor wants to arrange his books on a shelf. He has 30 books and only space on the shelf for 20 of them. How many different 20-book arrangements can he make using the 30 books This is an example of a problem that can be solved using which method
A) Permutations
B) Combinations
C) Conditional probability
D) Randomness

Accepted Answer

This problem involves selecting a subset of books (20 out of 30) where the order in which they are arranged on the shelf matters, making it a permutations problem.

Question 2

Solve Counting Problems Using Combinations
-A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 6 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen
A) 1680
B) 972
C) 261,273,600
D) 7,257,600

Accepted Answer

The number of ways to choose 6 multiple-choice questions from 9 is C(9,6) = 84. The number of ways to choose 3 open-ended problems from 6 is C(6,3) = 20. The total number of ways is 84 * 20 = 1680.

Question 3

Solve Counting Problems Using Combinations
-From 8 names on a ballot, a committee of 5 will be elected to attend a political national convention. How many different committees are possible
A) 56
B) 6720
C) 336
D) 3360

Accepted Answer

The answer of Solve Counting Problems Using Combinations
-From 8 names...

Question 4

Solve Counting Problems Using Combinations
-Professor Alle Whet teaches French and has a class of 24 students. Part of his grading system includes an observation of groups of 3 students engaged in a conversation in Fren
A) Combinations
B) Permutations
C) Conditional probability
D) Randomness

Accepted Answer

Combinations are used to solve this problem because the order in which the students are selected to form groups does not matter.

Question 5

Solve Counting Problems Using Combinations
-$${ } _ { 4 } \mathrm { C } _ { 3 }$$
A) 4
B) 3
C) 12
D) 24

Accepted Answer

$${ } _ { 4 } \mathrm { C } _ { 3 }$$ calculates the number of combinations of 4 items taken 3 at a time, which equals 4. This is because there are 4 ways to leave out one item from the set of 4.

Question 6

Solve Counting Problems Using Permutations
-How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if A and B must sit together
A) 48
B) 120
C) 12
D) 24

Accepted Answer

Since A and B must sit together, we can treat them as a single unit. This means we have 4 units to arrange (AB, C, D, E). The number of ways to arrange these 4 units is 4! = 24. However, within the AB unit, A and B can switch places, adding another factor of 2. Therefore, the total number of arrangements is 24 * 2 = 48.

Question 7

Solve Counting Problems Using Combinations
-${ } _ { 11 } \mathrm { C } _ { 5 }$
A) 462
B) 332,640
C) 27,720
D) 1440

Accepted Answer

${ } _ { 11 } \mathrm { C } _ { 5 } = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462.$

Question 8

Solve Counting Problems Using Permutations
-In a contest in which 9 contestants are entered, in how many ways can the 3 distinct prizes be awarded
A) 504
B) 60,480
C) 10,080
D) 120,960

Accepted Answer

The number of ways to award 3 distinct prizes among 9 contestants is calculated using permutations: 9P3 = 9! / (9-3)! = 9*8*7 = 504.

Question 9

Solve Counting Problems Using Permutations
-How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once
A) 90
B) 1,814,400
C) 45
D) 3,628,800

Accepted Answer

The word HYPERBOLAS consists of 10 distinct letters. The number of arrangements of 2 letters from these 10 is calculated using permutations: 10P2 = 10! / (10-2)! = 10 * 9 = 90.

Question 10

Solve Counting Problems Using Combinations
-To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 50 numbers (one through 50.) The order in which the selections is made does not matter. How many different selections are possible
A) 15,890,700
B) 13,983,816
C) 300
D) 720

Accepted Answer

The number of different selections is given by the combination formula:C(n, r) = n! / (n - r)! / r!where n is the total number of items and r is the number of items to be selected.In this case, n = 50 and r = 6, so the number of different selections is:C(50, 6) = 50! / (50 - 6)! / 6! = 15,890,700

Question 11

Solve Counting Problems Using Combinations
-$${ } _ { 6 } C _ { 1 }$$
A) 6
B) 720
C) 3
D) 240

Accepted Answer

$${ } _ { 6 } C _ { 1 }$$ calculates the number of ways to choose 1 item out of 6, which is simply 6.

Question 12

Solve Counting Problems Using Combinations
-${ } _ { 8 } C _ { 0 }$
A) 1
B) 8
C) 10,080
D) 20,160

Accepted Answer

The combination formula ${ } _ { n } C _ { r } = \frac{n!}{r!(n-r)!}$ simplifies to 1 when $r = 0$, regardless of the value of $n$, because $0! = 1$.

Question 13

Solve Counting Problems Using Combinations
-In how many ways can a committee of three men and four women be formed from a group of 10 men and 10 women
A) 25,200
B) 3,628,800
C) 50,400
D) 70

Accepted Answer

First, select 3 men from 10, which is 10C3 = 120 ways. Then, select 4 women from 10, which is 10C4 = 210 ways. Multiply these together for total combinations: 120 * 210 = 25,200.

Question 14

Solve Counting Problems Using Permutations
-The Environmental Protection Agency must inspect nine factories for complaints of water pollution. In how many different ways can a representative visit five of these to investigate this week
A) 15,120
B) 362,880
C) 5
D) 45

Accepted Answer

This is a permutation problem because the order in which the factories are visited matters. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items to choose from, and r is the number of items to choose. Here, n=9 (factories) and r=5 (factories to visit). So, P(9, 5) = 9! / (9-5)! = 9! / 4! = (9*8*7*6*5*4*3*2*1) / (4*3*2*1) = 15,120.

Question 15

Solve Counting Problems Using Permutations
-How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if D and E will not sit next to each other
A) 72
B) 24
C) 48
D) 60

Accepted Answer

First, calculate the total number of ways all five people can sit in a row without restrictions, which is 5! = 120. Then, calculate the number of ways D and E can sit next to each other by treating them as a single unit, giving us 4! ways to arrange the units, and 2! ways to arrange D and E within their unit, for a total of 4! * 2! = 48 ways. Subtracting these from the total gives 120 - 48 = 72 ways where D and E do not sit next to each other.

Question 16

Solve Counting Problems Using Combinations
-$10 { } ^ { C _ { 10 } }$
A) 1
B) $3,628,800$
C) 907,200
D) $0.5$

Accepted Answer

The notation $10 { } ^ { C _ { 10 } }$ represents the number of ways to choose 10 items out of 10, which is always 1, since there is only one way to choose all items from a set of items.

Question 17

Solve Counting Problems Using Permutations
-In how many different ways can a ski club consisting of 20 people select a person for its officers The positions available are president, vice president, treasurer, and secretary. No person can hold more than one positions and the each office is filled in order.
A) 116,280
B) 4845
C) 1440
D) 74

Accepted Answer

To find the number of different ways the ski club can select its officers, we need to use the permutation formula: nPr = n!/(n-r)!, where n is the number of people and r is the number of positions to be filled.

For the first position of president, any one of the 20 people can be selected. Thus, there are 20 options.
For the second position of vice president, after the president has been selected, there are only 19 people remaining who can be selected. Thus, there are 19 options.
Similarly, for the third and fourth positions of treasurer and secretary, there are only 18 and 17 people respectively who can be selected.

Using the permutation formula, we get:
nPr = 20P4
= 20!/(20-4)!
= 20!/16!
= (20x19x18x17x16!)/16!
= 20x19x18x17
= 116,280

Therefore, the best choice is A, with the explanation that we used the permutation formula to find the number of different ways the ski club can select its officers, with 20 options for the first position, 19 options for the second position, 18 options for the third position, and 17 options for the fourth position. This results in a total of 116,280 different ways to select the officers.

Question 18

Solve Counting Problems Using Permutations
-How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if C must sit to the right of but not necessarily next to B
A) 60
B) 24
C) 20
D) 48

Accepted Answer

Treat B and C as a single entity since C must be to the right of B, giving us 4 entities to arrange (BC, A, D, E). There are 4! ways to arrange these, which is 24. However, within the BC entity, B and C can switch places, doubling the number of arrangements to 48. But since C must always be to the right of B, we only consider the arrangements where B is before C, which are already counted in the 24 arrangements. Therefore, we must also account for the different ways the remaining three people can be arranged, which is already considered in the 4! calculation. Thus, the correct calculation should consider both the arrangements of the 4 entities and the internal arrangement of B and C, leading to 4! = 24 ways to arrange the entities, and since B and C can be considered as a single entity with a fixed order (B before C), we don't multiply by additional permutations for their arrangement. The initial explanation mistakenly suggested multiplying by 2 for the BC arrangement, which is incorrect since C's position relative to B is fixed. The correct number of total arrangements is 4! = 24 for the entities, considering the constraint on B and C's positions.

Question 19

Solve Counting Problems Using Combinations
-$\frac { { } _ { 9 } C _ { 3 } } { { } _ { 6 } C _ { 4 } }$
A) $\frac { 28 } { 5 }$
B) $\frac { 7 } { 5 }$
C) $\frac { 14 } { 5 }$
D) 4032

Accepted Answer

The answer of Solve Counting Problems Using Combinations
-\(\frac { {...

Question 20

Solve Counting Problems Involving Permutations with Nondistinct Items
-How many distinct arrangements can be formed from all the letters of students
A) 10,080
B) 1680
C) 720
D) 40,320

Accepted Answer

The word "students" has 8 letters with "t" and "s" each appearing twice. The number of distinct arrangements is calculated using the formula for permutations of a multiset: n! / (n1! * n2! * ... * nk!), where n is the total number of items, and n1, n2, ..., nk are the counts of each distinct item. Here, it's 8! / (2! * 2!) = 40,320 / (2 * 2) = 10,080.

Quiz 5: Probability

Solve Counting Problems Using Combinations -A professor wants to arrange his books on a shelf. He has 30 books and only space on the shelf for 20 of them. How many different 20-book arrangements can he make using the 30 books This is an example of a problem that can be solved using which method

Solve Counting Problems Using Combinations -From 8 names on a ballot, a committee of 5 will be elected to attend a political national convention. How many different committees are possible

Solve Counting Problems Using Combinations -Professor Alle Whet teaches French and has a class of 24 students. Part of his grading system includes an observation of groups of 3 students engaged in a conversation in Fren

Solve Counting Problems Using Combinations - ${ } _ { 4 } \mathrm { C } _ { 3 }$

Solve Counting Problems Using Permutations -How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if A and B must sit together

Solve Counting Problems Using Combinations - ${ } _ { 11 } \mathrm { C } _ { 5 }$

Solve Counting Problems Using Permutations -In a contest in which 9 contestants are entered, in how many ways can the 3 distinct prizes be awarded

Solve Counting Problems Using Permutations -How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once

Solve Counting Problems Using Combinations -To win at LOTTO in a certain state, one must correctly select 6 numbers from a collection of 50 numbers (one through 50.) The order in which the selections is made does not matter. How many different selections are possible

Solve Counting Problems Using Combinations - ${ } _ { 6 } C _ { 1 }$

Solve Counting Problems Using Combinations - ${ } _ { 8 } C _ { 0 }$

Solve Counting Problems Using Combinations -In how many ways can a committee of three men and four women be formed from a group of 10 men and 10 women

Solve Counting Problems Using Permutations -The Environmental Protection Agency must inspect nine factories for complaints of water pollution. In how many different ways can a representative visit five of these to investigate this week

Solve Counting Problems Using Permutations -How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if D and E will not sit next to each other

Solve Counting Problems Using Combinations - $10 { } ^ { C _ { 10 } }$

Solve Counting Problems Using Permutations -How many ways can five people, A, B, C, D, and E, sit in a row at a concert hall if C must sit to the right of but not necessarily next to B

Solve Counting Problems Using Combinations - $\frac { { } _ { 9 } C _ { 3 } } { { } _ { 6 } C _ { 4 } }$

Solve Counting Problems Involving Permutations with Nondistinct Items -How many distinct arrangements can be formed from all the letters of students