Question 1

Distinguish Between Permutation Problems and Combination Problems
-From 9 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?
A) permutations
B) combinations

Accepted Answer

This is a combination problem because the order in which the committee members are chosen does not matter.

Question 2

Use the Permutations Formula
-The matching section of an exam has 5 questions and 7 possible answers. In how many different ways can a student answer the 5 questions, if none of the answer choices can be repeated?
A) 2520
B) 42
C) 21
D) 84

Accepted Answer

This is a permutation problem because the order in which the answers are matched to the questions matters and no answer can be used more than once. The formula for permutations when choosing r items from a set of n items is nPr = n! / (n-r)!. Here, n = 7 (possible answers) and r = 5 (questions). So, 7P5 = 7! / (7-5)! = 7! / 2! = (7*6*5*4*3) / (2*1) = 2520.

Question 3

Use the Combinations Formula
-A hamburger shop sells hamburgers with cheese, relish, lettuce, tomato, onion, mustard, or ketchup. How many different hamburgers can be concocted using any 3 of the extras?
A) 35
B) 210
C) 840
D) 105

Accepted Answer

The combinations formula is used to find the number of ways to choose k items from n items without regard to the order of selection. The formula is C(n, k) = n! / [k!(n-k)!]. Here, n = 7 (cheese, relish, lettuce, tomato, onion, mustard, ketchup) and k = 3. Thus, C(7, 3) = 7! / [3!(7-3)!] = 7! / (3!4!) = (7×6×5) / (3×2×1) = 35.

Question 4

Use the Combinations Formula
-Ron finds 8 books at a bookstore that he would like to buy, but he can afford only 5 of them. In how many ways can he make his selection? How many ways can he make his selection if he decides that one of the books is a must?
A) 56; 35
B) 6720; 840
C) 336; 210
D) 3360; 420

Accepted Answer

The number of ways to choose 5 books out of 8 is calculated using the combinations formula, which is C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. So, for choosing 5 out of 8, it's C(8, 5) = 8! / [5!(8-5)!] = 56. If one book is a must, then he is choosing 4 out of the remaining 7, which is C(7, 4) = 7! / [4!(7-4)!] = 35.

Question 5

Additional Concepts
-$$\frac { 9 ^ { P _ { 3 } } } { 3 ! } - 9 C _ { 3 }$$
A) 0
B) 6
C) 362,880
D) 720

Accepted Answer

The expression simplifies to 0 because $9P_3$ represents the number of permutations of 9 items taken 3 at a time, which is $\frac{9!}{(9-3)!} = \frac{9!}{6!}$, and $9C_3$ represents the number of combinations of 9 items taken 3 at a time, which is $\frac{9!}{3!6!}$. When you simplify the given expression, both terms actually represent the same value, leading to their difference being 0.

Question 6

Use the Permutations Formula
-A church has 9 bells in its bell tower. Before each church service 5 bells are rung in sequence. No bell is rung more than once. How many sequences are there?
A) 15,120
B) 3024
C) 126
D) 6048

Accepted Answer

The number of sequences is calculated using the permutations formula, which is P(n, r) = n! / (n-r)!, where n is the total number of items, and r is the number of items to be arranged. Here, n=9 (bells) and r=5 (bells to be rung in sequence), so P(9, 5) = 9! / (9-5)! = 9! / 4! = 15,120.

Question 7

Use the Combinations Formula
-From 9 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?
A) 126
B) 3024
C) 15,120
D) 1512

Accepted Answer

The number of ways to choose 4 names out of 9 is calculated using the combinations formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$. Here, $n=9$ and $k=4$, so $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = 126$.

Question 8

Distinguish Between Permutation Problems and Combination Problems
-A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 11 members and any member can be elected to each position? No person can hold more than one office.
A) permutations
B) combinations

Accepted Answer

This is a permutation problem because the order in which the officers are elected matters (president, vice-president, and secretary-treasurer are distinct positions).

Question 9

Use the Combinations Formula
-In a student government election, 6 seniors, 2 juniors, and 3 sophomores are running for election. Students elect four at-large senators. In how many ways can this be done?
A) 330
B) 7920
C) 1,663,200
D) 36

Accepted Answer

The total number of candidates is 6 seniors + 2 juniors + 3 sophomores = 11 candidates. The number of ways to choose 4 out of 11 candidates, regardless of class year, is calculated using the combinations formula, which is C(n, k) = n! / [k!(n - k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. Thus, C(11, 4) = 11! / [4!(11 - 4)!] = 11! / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330.

Question 10

Use the Combinations Formula
-A stack of 9 different cards are shuffled and spread out face down. If 4 cards are turned face up, how many different 4-card combinations are possible?
A) 126
B) 3024
C) 15,120
D) 1512

Accepted Answer

The number of different 4-card combinations possible from a stack of 9 different cards is calculated using the combinations formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items, and $k$ is the number of items to choose. Here, $n=9$ and $k=4$, so $C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = 126$.

Question 11

Use the Permutations Formula
-How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0? No digit can be used more than once.
A) 720
B) 604,800
C) 120
D) 1,209,600

Accepted Answer

The number of 3-digit numbers that can be formed from 10 digits (0-9) without repetition, where the first digit can't be 0 (9 options), and then 9 options for the second digit, and 8 for the third, is calculated by 9 * 9 * 8 = 720.

Question 12

Distinguish Between Permutation Problems and Combination Problems
-The matching section of an exam has 4 questions and 9 possible answers. In how many different ways can a student answer the 4 questions, if none of the answer choices can be repeated?
A) permutations
B) combinations

Accepted Answer

This is a permutation problem because the order in which the answers are matched to the questions matters.

Question 13

Use the Permutations Formula
-How many 2-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once.
A) 30
B) 360
C) 15
D) 720

Accepted Answer

The number of 2-letter codes that can be formed from 6 letters without repetition is calculated using the permutation formula P(n, r) = n! / (n-r)!, where n is the total number of items to choose from, r is the number of items to choose, n! denotes the factorial of n, and (n-r)! is the factorial of the difference between n and r. Here, n=6 (letters A, B, C, D, E, F) and r=2 (because we are forming 2-letter codes), so P(6, 2) = 6! / (6-2)! = 6! / 4! = (6×5×4!)/4! = 30.

Question 14

Distinguish Between Permutation Problems and Combination Problems
-In a student government election, 5 seniors, 2 juniors, and 3 sophomores are running for election. Students elect four at-large senators. In how many ways can this be done?
A) permutations
B) combinations

Accepted Answer

This is a combination problem because the order in which the four senators are elected does not matter. What matters is which four students are chosen out of the total, regardless of the order.

Question 15

Use the Permutations Formula
-In how many ways can Susan arrange 10 books into 4 slots on her bookshelf?
A) 5040
B) 151,200
C) 25,200
D) 302,400

Accepted Answer

There are 10 books and 4 slots, so the number of ways to arrange the books is 10P4 = 5040.

Question 16

Use the Combinations Formula
-8C0
A) 1
B) 80
C) 0
D) 40,320 Solve the problem.

Accepted Answer

The combinations formula is given by nCr = n! / r!(n-r)!, where n is the total number of items, and r is the number of items to choose. For 8C0, it means choosing 0 items out of 8, which is always equal to 1 because there is exactly one way to choose nothing.

Question 17

Use the Combinations Formula
-11C3
A) 165
B) 6,652,800
C) 13,305,600
D) 990

Accepted Answer

The combinations formula is given by $nCk = \frac{n!}{k!(n-k)!}$. For 11C3, it is $\frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$.

Question 18

Use the Permutations Formula
-A combination lock has 20 numbers on it. How many different 3-digit lock combinations are possible if no digit can be repeated?
A) 6840
B) 2280
C) 1140
D) 380

Accepted Answer

The number of different 3-digit lock combinations possible without repeating any digit can be calculated using the permutations formula, which is n! / (n - r)!, where n is the total number of options (20 numbers) and r is the number of selections to make (3 digits). So, it's 20! / (20 - 3)! = 20! / 17! = 20 x 19 x 18 = 6840.

Question 19

Additional Concepts
-$1 - \frac { { } _ { 9 } \mathrm { P } _ { 6 } } { { } _ { 12 } \mathrm { P } _ { 9 } }$
A) $\frac { 1319 } { 1320 }$
B) $\frac { 1 } { 1320 }$
C) $\frac { 34 } { 55 }$
D) $\frac { 1321 } { 1320 }$

Accepted Answer

The expression simplifies to $1 - \frac{9!/(9-6)!}{12!/(12-9)!}$, which equals $1 - \frac{60480}{7983360} = 1 - \frac{1}{132}$. Simplifying gives $\frac{1319}{1320}$.

Question 20

Use the Permutations Formula
-A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 9 members and any member can be elected to each position? No person can hold more than one office.
A) 504
B) 252
C) 168
D) 3024

Accepted Answer

The number of ways to elect a president, vice-president, and secretary-treasurer from 9 members, with no repetition, is calculated using permutations: 9P3 = 9! / (9-3)! = 9! / 6! = 9 × 8 × 7 = 504.

Quiz 11: Sequences, Induction, and Probability

Distinguish Between Permutation Problems and Combination Problems -From 9 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?

Use the Permutations Formula -The matching section of an exam has 5 questions and 7 possible answers. In how many different ways can a student answer the 5 questions, if none of the answer choices can be repeated?

Use the Combinations Formula -A hamburger shop sells hamburgers with cheese, relish, lettuce, tomato, onion, mustard, or ketchup. How many different hamburgers can be concocted using any 3 of the extras?

Use the Combinations Formula -Ron finds 8 books at a bookstore that he would like to buy, but he can afford only 5 of them. In how many ways can he make his selection? How many ways can he make his selection if he decides that one of the books is a must?

Additional Concepts - $\frac { 9 ^ { P _ { 3 } } } { 3 ! } - 9 C _ { 3 }$

Use the Permutations Formula -A church has 9 bells in its bell tower. Before each church service 5 bells are rung in sequence. No bell is rung more than once. How many sequences are there?

Use the Combinations Formula -From 9 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?

Distinguish Between Permutation Problems and Combination Problems -A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 11 members and any member can be elected to each position? No person can hold more than one office.

Use the Combinations Formula -In a student government election, 6 seniors, 2 juniors, and 3 sophomores are running for election. Students elect four at-large senators. In how many ways can this be done?

Use the Combinations Formula -A stack of 9 different cards are shuffled and spread out face down. If 4 cards are turned face up, how many different 4-card combinations are possible?

Use the Permutations Formula -How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0? No digit can be used more than once.

Distinguish Between Permutation Problems and Combination Problems -The matching section of an exam has 4 questions and 9 possible answers. In how many different ways can a student answer the 4 questions, if none of the answer choices can be repeated?

Use the Permutations Formula -How many 2-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once.

Distinguish Between Permutation Problems and Combination Problems -In a student government election, 5 seniors, 2 juniors, and 3 sophomores are running for election. Students elect four at-large senators. In how many ways can this be done?

Use the Permutations Formula -In how many ways can Susan arrange 10 books into 4 slots on her bookshelf?

Use the Combinations Formula -8C0

Use the Combinations Formula -11C3

Use the Permutations Formula -A combination lock has 20 numbers on it. How many different 3-digit lock combinations are possible if no digit can be repeated?

Additional Concepts - $1 - \frac { { } _ { 9 } \mathrm { P } _ { 6 } } { { } _ { 12 } \mathrm { P } _ { 9 } }$

Use the Permutations Formula -A club elects a president, vice-president, and secretary-treasurer. How many sets of officers are possible if there are 9 members and any member can be elected to each position? No person can hold more than one office.