5.2 Counting Continued
How many 10-bit strings are there subject to each of the following restrictions? (a) No restrictions. (b) The string starts with 001. (c) The string starts with 001 or 10. (d) The 1st 2 bits are the same as the last 2 bits.
(a) 2 choices for each of the 10 bits: 2¹⁰ (b) The first 3 bits of the string are determined. Each of the remaining 7 bits can be 0 or 1. Therefore 2⁷ binary strings of length 10 start with 001. (c) There are 2⁷ 10-bit strings that begin with 001. There are 2⁸ 10-bit strings that begin with 10. A string can't begin with 001 and 10, so the sum rule applies. Therefore the number of 10-bit strings that begin with 001 or 10 is 2⁷ + 2⁸. (d) Each of the first 8 bits can be either 0 or 1. Once the first 8 bits are determined, the last two bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is 2⁸
A group of five friends go to a restaurant for dinner. The restaurant offers 20 different main dishes. (a) Suppose that the group collectively orders five different dishes to share. The waiter just needs to place all five dishes in the center of the table. How many different possible orders are there for the group? (b) Suppose that each individual orders a main course. The waiter must remember who ordered which dish as part of the order. It's possible for more than one person to order the same dish. How many different possible orders are there for the group? (c) Suppose that each individual orders a main course. The waiter must remember who ordered which dish as part of the order. However the friends agree that no two people will order the same dish.
(a) C(20, 5). The order is just a subset of five of the 20 dishes on the menu. (b) 20⁵ Each person can select any of the 20 dishes on the menu. There are five people each of whom is selecting a dish. The 20 choices for each person are combined using the product rule. (c) P(20,5). Each person selects a dish in turn. The order in which the dishes are selected matters because that determines who gets which dish. Since no dish is selected more than once, the entire order is a sequence of 5 out of the 20 dishes with no repetitions.
A country has two political parties, the Demonstrators and the Repudiators. Suppose that the national senate consists of 100 members, 44 of which are Demonstrators and 56 of which are Repudiators. (a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators? (b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?
(a) C(44, 5)* C(56, 5) The committee must have 5 Demonstrators and 5 Repudiators. There are C(44, 5) ways to select the 5 committee members from the 44 Demonstrators. There are C(56, 5) ways to select the 5 committee members from the 56 Repudiators. Therefore there are C(44, 5)* C(56, 5) ways to select the entire committee. (b) P(44,2) · P(56,2) - There are 44 ways to select the speaker from the Demonstrators. Once that person is chosen, there are 43 ways to select the vice speaker from the remaining Demonstrators. Therefore, there are P(44, 2) = 44 · 43 ways to select the speaker and vice speaker from the Demonstrators. Similarly, there are P(56, 2) ways to select the speaker and vice speaker from the Repudiators. The choices are combined using the product rule because a speaker and vice speaker are selected from both parties. Therefore there are a total of P(44,2) · P(56,2) ways to select the speaker and vice speaker from the two parties.
(a) How many different five-card hands are there from a standard deck of 52 playing cards? (b) How many five-card hands have exactly two hearts? (c) How many five-card hands are made entirely of hearts and diamonds? (d) How many five-card hands have four cards of the same rank?
(a) C(52, 5) (b) There are C(13, 2) ways to select a subset of 2 hearts from the set of 13 hearts in the deck. There are C(39, 3) ways to select a subset of 3 non-hearts from the set of 39 non-hearts in the deck. Since the hearts and the non-hearts must all be selected in order to determine the cards in the hand, the product rule applies, and there are a total of C(13, 2) * C(39,3) ways to select a 5-card hand with exactly two hearts. (c) There are C(26, 5) ways to select a subset of 5 cards from the set of 26 (13 hearts and 13 diamonds) in the deck. (d) There are 13 ways to select a rank. Once a rank has been chosen, all four of the cards with that rank will be in the hand. Then there are 48 ways to select the remaining card that does not have the same rank as the four already chosen. The total number of ways to select a five-card hand with four cards of the same rank is 13·48.
20 different comic books will be distributed to five kids. (a) How many ways are there to distribute the comic books if there are no restrictions on how many go to each kid (other than the fact that all 20 will be given out)? (b) How many ways are there to distribute the comic books if they are divided evenly so that 4 go to each kid?
(a) For each comic book, there are five different choices for which kid will receive that comic book. 20 decisions are made, one for each comic book with five different possibilities for each choice. The number of ways to distribute the comic books to the five kids is 5²⁰ (b) There are C(20, 4) ways to select the comic books given to the first kid. Then C(16, 4) ways to select the comic books given to the second kid, then C(12, 4) for the third kid, and C(8, 4) for the fourth kid. The fifth kid gets the four remaining comic books. Putting together the selections by the product rule, the number of ways to distribute the comic books to the kids with four going to each kid is (see picture)
A school cook plans her calendar for the month of February in which there are 20 school days. She plans exactly one meal per school day. Unfortunately, she only knows how to cook ten different meals. (a) How many ways are there for her to plan her schedule of menus for the 20 school days if there are no restrictions on the number of times she cooks a particular type of meal? (b) How many ways are there for her to plan her schedule of menus if she wants to cook each meal the same number of times?
(a) For each day, there are ten different choices for what she will cook that day. 20 decisions are made, one for each day with ten different possibilities for each choice. The number of ways to plan her schedule of menus for the 20 school days is 10²⁰. (b) If she cooks each meal the same number of times, then she will cook each meal twice. Number the meals 1 through 10. For j = 1 to 10, she selects the days in which she will cook meal j. After she has selected the days for meals 1 through j, there are 20 - 2j days remaining from which to select the two days for meal j+1. The total number of choices is shown as follows:
A 5-card hand is drawn from a deck of standard playing cards. (a) How many 5-card hands have at least one club? (b) How many 5-card hands have at least two cards with the same rank?
(a) The total number of 5-card hands with no restrictions is C(52, 5). The total number of 5-card hands with no clubs is C(39, 5) because there are 39 non-clubs (13 hearts + 13 spades + 13 diamonds) from which to select the five cards. Therefore the number of 5-card hands with at least one club is C(52, 5) − C(39, 5). (b) The total number of 5-card hands with no restrictions is C(52, 5). The number of 5-card hands in which all the cards have different ranks is 4⁵⋅C(13, 5). Therefore the number of 5-card hands with at least two cards of the same rank is C(52, 5) − 4⁵⋅C(13, 5).
Ten members of a club are lining up in a row for a photograph. The club has one president and one VP. (a) How many ways are there for the club members to line up in which the president is not next to the VP? (b) How many ways are there for the club members to line up if the VP is not in the leftmost position? (c) How many ways are there for the club members to line up if the VP is not at one end (i.e. in the leftmost or rightmost positions)?
(a) There are 10! ways to line up the ten people in the club if there are no restrictions on how the people are placed. As shown earlier, the number of ways to line up the ten people in the club in which the president and VP are next to each other is 2·9!. Therefore the number of ways to line up the ten people in the club in which the president is not next to the VP is 10! - 2·9! = 8·9!. (b) There are 10! ways to line up the ten people in the club if there are no restrictions on how the people are placed. The number of ways to line up the ten people if the VP is in the leftmost position is 9!. (First place the VP in the leftmost position. Then there are 9! ways to order the rest of the people.) Therefore the number of ways to line up the ten people in the club in which the VP is not in the leftmost position is 10! - 9! = 9·9!. (c) There are 10! ways to line up the ten people in the club if there are no restrictions on how the people are placed. The number of ways to line up the ten people if the VP is in the leftmost position is 9!. The number of ways to line up the ten people if the VP is in the rightmost position is 9!. The number of ways to line up the ten people if the VP is in the rightmost or the leftmost position is 9! + 9! = 2·9!. Therefore the number of ways to line up the ten people in the club in which the VP is not in the leftmost or rightmost position is 10! - 2·9! = 8·9!.
A search committee is formed to find a new software engineer. (a) If 100 applicants apply for the job, how many ways are there to select a subset of 9 for a short list? (b) If 6 of the 9 are selected for an interview, how many ways are there to pick the set of people who are interviewed? (You can assume that the short list is already decided). (c) Based on the interview, the committee will rank the top three candidates and submit the list to their boss who will make the final decision. (You can assume that the interviewees are already decided.) How many ways are there to select the list from the 6 interviewees?
(a) There are C(100, 9) ways to select a subset of 9 people from a set of 100. (b) There are C(9, 6) ways to select a subset of 6 people from a set of 9. (c) P(6, 3). The selected list is ordered because the three people selected are also designated as first choice, second choice, and third choice. The number of ways to select an ordered list of three people without repetitions from a set of 6 people is P(6, 3).
How many different strings of length 12 containing exactly five a's can be chosen over the following alphabets? (a) The alphabet {a, b} (b) The alphabet {a, b, c}
(a) There are C(12, 5) ways to select where the 5 a's will be placed among the 12 possible locations. Once the a's are placed, the unfilled locations are filled with b's. Therefore the number of strings over the alphabet {a, b} with exactly five a's is C(12, 5). (b) There are C(12, 5) ways to select where the five a's will be placed among the 12 possible locations. Once the a's are placed, there are two choices (b or c) for each of the remaining seven unfilled locations. Therefore, there are 2⁷ ways to fill the locations that do not have an a. The number of strings over the alphabet {a, b, c} with exactly five a's is 2⁷⋅C(12,5).
14 students have volunteered for a committee. 8 of them are seniors and 6 of them are juniors. (a) How many ways are there to select a committee of 5 students? (b) How many ways are there to select a committee with 3 seniors and 2 juniors? (c) Suppose the committee must have five students (either juniors or seniors) and that one of the five must be selected as chair. How many ways are there to make the selection?
(a) There are C(14, 5) ways to select a subset of 5 students from a set of 14 students. (b) There are C(8, 3) ways to select a subset of 3 seniors from a set of 8 seniors. There are C(6, 2) ways to select a subset of 2 juniors from a set of 6 juniors. Since the juniors and the seniors must all be selected in order to determine the membership of the committee, the product rule applies, and there are a total of C(8, 3) * C(6, 2) ways to make the selection. (c) There are 14 ways to select the chair from among the 14 students. Once the chair has been chosen, there are C(13, 4) ways to select the remaining four members of the committee from the 13 students who were not selected already. The total number of ways to select a committee of 5 in which one of the 5 is designated as the chair is 14⋅C(13, 4).
Suppose a network has 40 computers of which 5 fail. (a) How many possibilities are there for the five that fail? (b) Suppose that 3 of the computers in the network have a copy of a particular file. How many sets of failures wipe out all the copies of the file? That is, how many 5-subsets contain the three computers that have the file?
(a) There are C(40, 5) ways to select a subset of 5 computers from a set of 40 computers. (b) Any such 5-subset of computers that fail must contain the 3 computers that have the file. The two remaining computers in the subset must be selected from the 37 computers that did not contain the file. Thus, the number of such 5-subsets in which the 3 computers with the file fail is the number of ways to select 2 computers from the 37 computers without the file, or C(37, 2).
How many ways are there to deal hands from a standard playing deck to four players if: (a) Each player gets exactly 13 cards. (b) Each player gets seven cards and the rest of the cards remain in the deck?
(a) There are C(52, 13) ways to pick 13 cards for the first player. Then C(39, 13) ways to pick 13 cards from the remaining 39 cards for the second player. Then C(26, 13) ways to pick 13 cards from the remaining 26 cards for the third player. The fourth player is left with the remaining 13 cards. Putting together the selections by the product rule, the number of ways to deal hands with 13 cards each to 4 players is: (b) There are C(52, 7) ways to pick 7 cards for the first player. Then C(45, 7) ways to pick 7 cards from the remaining 45 cards for the second player. Then C(38, 7) ways to pick 7 cards from the remaining 38 cards for the third player. Then C(31, 7) ways to pick 7 cards from the remaining 31 cards for the fourth player. The 24 remaining cards remain in the deck. Putting together the selections by the product rule, the number of ways to deal hands with 7 cards each to 4 players is:
a) How many 8-bit strings have at least two consecutive 0's or two consecutive 1's? (b) How many 8-bit strings do not begin with 000?
(a) if a string does not have two consecutive 0's or two consecutive 1's, then the bits alternate. There are only two 8-bit strings in which the bits alternate: 10101010 and 01010101. The total number of 8-bit strings with no restrictions is 2⁸ = 256. Therefore the number of 8-bit strings which have at least two consecutive 0's or at least two consecutive 1's is 2⁸ - 2 → 256-2 = 254 (b) The number of 8-bit strings that do begin with 000 is 2⁵ because there are two choices for each of the last five bits. The total number of 8-bit strings with no restrictions is 2⁸. Therefore the number of 8-bit strings which do not begin with 000 is 2⁸ - 2⁵.
Consider a function f that maps 5-permutations from the set S = {1, ..., 20} to 5-subsets from S. The function takes a 5-permutation and creates an unordered set whose elements are the five numbers included in the permutation. (a) What is the value of f on input (12, 1, 3, 15, 9)? (b) Is (12, 3, 12, 4, 19) a 5-permutation? Why or why not? (c) How many permutations are mapped onto the subset {12, 3, 13, 4, 19}?
(a) {1, 3, 9, 12, 15} (b) No because 12 appears twice. (c) 5! because there are 5! ways to order the elements in {12, 3, 13, 4, 19}.
(e) A "full house" is a five-card hand that has two cards of the same rank and three cards of the same rank. For example, {queen of hearts, queen of spades, 8 of diamonds, 8 of spades, 8 of clubs}. How many five-card hands contain a full house? (f) How many five-card hands do not have any two cards of the same rank?
(e) There are 13 ways to select the rank for the two cards with the same rank. The three cards with the same rank must have a different rank than the rank chosen for the pair, so there are 12 ways to select the rank for the three cards with the same rank. Once the rank has been chosen for the two cards that have the same rank, there are C(4, 2) ways to select two cards from the four cards with that rank. Once the rank has been chosen for the three cards that have the same rank, there are C(4, 3) ways to select three cards from the four cards with that rank. The choices are put together by the product rule, so that the number of ways to select a hand that is full house is: 13*12*C(4, 2)*C(4, 3) (f) There are 13 different possible ranks. The number of ways to select 5 distinct ranks from 13 possible ranks is C(13, 5). For each rank chosen, there are four possible cards with that rank that can be selected. Therefore once the ranks have been determined, there are 4⁵ ways to select the cards in the hand. The total number of ways to select a five-card hand in which no two cards have the same rank is 4⁵⋅C(13, 5).
How many 10-bit strings are there subject to each of the following restrictions? (e) The string has exactly six 0's. (f) The string has exactly six 0's and the first bit is 1. (g) There is exactly one 1 in the first half and exactly three 1's in the second half.
(e) There are C(10, 6) ways to select where the six 0's will be placed among the ten possible locations. Once the 0's are placed, the four 1's go in the unfilled locations. Therefore the number of 10-bit strings with exactly six 0's is C(10, 6). (f) The six 0's can be placed in any of the 10 locations, except the first location. There are C(9, 6) ways to select the six locations for the 0's among the last nine locations in the string. Once the 0's are placed, the four 1's go in the unfilled locations. Therefore the number of 10-bit strings with exactly six 0's and a 1 in the first location is C(9, 6). (g) There are C(5, 1) ways to select the location for the 1 among the first five locations. There are C(5, 3) ways to select the location for the three 1's among the last five locations. Once the 1's are placed, the 0's go in all the unfilled locations. Since the location for the 1 in the first half and the locations for the 1's in the second half must both be determined to specify the string, the product rule applies. Therefore the number of 10-bit strings with one 1 in the first half and three 1's in the second half is C(5, 1) C(5, 3)= 5*C(5,3).
Calculate a numerical value for C(100, 100), the number of ways to select a subset of size 100 from a set of size 100. Note that 0! = 1.
1
Calculate a numerical value for C(100, 1), the number of ways to select a subset of size 1 from a set of size 100.
100
Define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and the 8-bit string with exactly five 1's - 11111111. A subset X of S with five elements maps on to a string x so that j ∈ X if and only if the j^th bit of x is 1. What string corresponds to the set {1, 3, 4, 5, 8}?
10111001 12345678 - {1, 3, 4, 5, 8} The bit locations of the string are numbered 1 through 8. There should be 1's in locations 1, 3, 4, 5, and 8 and 0's in the other locations.
A teacher must select four members of the math club to participate in an upcoming competition. How many ways are there for her to make her selection if the club has 12 members?
12 choose 4 She is selecting a 4-subset from the 12 kids in the club
A file will be replicated on 3 different computers in a distributed network of 15 computers. How many ways are there to select the locations for the file?
15 choose 3 In determining the locations for the file, a 3-subset is selected from the set of 15 computers
Let S = {a, b, c}. Is (b, a) a 2-permutation or a 2-subset from S?
2-permutation The regular parens () indicate that the order of the elements matters, so (b, a) ≠ (a, b).
Let S = {a, b, c}. Is {b, a} a 2-permutation or a 2-subset from S?
2-subset The curly braces {} indicates that order does not matter, so {b, a} = {a, b}.
A bit string contains 1's and 0's. How many different bit strings can be constructed given the restriction(s)? Length is 25. Starts with: 101
25 - 3 positions remain, each with 2 possibilities: 2²²
Let S = {a, b, c}. How many different 2-permutations from S are there?
3! = 6 The 2-permutations from S are: (a, b),(b, a),(a, c),(c, a),(b, c),(c, b).
A class of 30 students elects four students to serve on a student leadership council. The teacher tallies the votes and only reveals the names of the four students who received the most votes. How many different outcomes are there from the election process?
30 choose 4 The students only learn the names of the four students who are elected. Each student that is elected plays the same role as the others. Therefore, the outcome is just a subset of four students chosen from the class of 30 students.
Calculate a numerical value for C(7, 3)
35
Calculate a numerical value for C(7, 4)
35
Dave swims three times in the week. How many ways are there to plan his workout schedule (i.e. which days he will swim) for a given week?
C(7, 3) 7 choose 3 - a subset. A schedule for Dave's workouts consists of the set of three days he will swim (e.g., {Mon, Wed, Sat}). The order in which he selects the three days of the week does not affect the schedule.Therefore, Dave is selecting a subset of 3 days from the 7 days of the week.
An auto dealer has 4 different cars and 6 different trucks.How many ways are there to select two vehicles?
Choose 2 cars from 10 total options: C(10, 2)
A shop has 4 different shirts and 5 different jeans.How many ways are there to select 2 shirts?
Choose 2 from 4 shirts: C(4, 2)
A bit string contains 1's and 0's. How many different bit strings can be constructed given the restriction(s)? Length is 20. Has exactly seven 0's.
Choose seven locations for 0's from 20 locations: C(20, 7)
A bit string contains 1's and 0's. How many different bit strings can be constructed when length is 18?
Each position has 2 possibilities: 2¹⁸
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
First select the player who will play center. There are only 3 possible choices for the center. After the center has been selected, the coach can place any of the remaining 11 players in any of the other four positions, so there are P(11, 4) choices for the players who will play the positions besides center. The total number of choices is 3·P(11, 4).
A bit string contains 1's and 0's. How many different bit strings can be constructed given the restriction(s)? Length is 28. Has exactly six 0's. Starts with: 11
First two locations already chosen.Choose six locations for 0's from 26 locations: C(26, 6)
A bit string contains 1's and 0's. How many different bit strings can be constructed given the restriction(s)? Length is 15.Has exactly twelve 0's.Starts with: 11
First two locations already chosen.Choose twelve locations for 0's from 13 locations: C(13, 12)
A shop has 4 different shirts and 8 different jeans. How many ways are there to select two items so that at least one jeans is chosen?
First, choose 2 from all items: C(12, 2) = 66 Second, choose all # of choices that make up no jeans. This way you can subtract this number from the total # of choices that do not include jeans at all. This is count by complement and guarantees all remaining answers will have at least one jean. Choose 2 from 4 total shirts: C(4, 2) = 6 Finally, subtract only shirts from all item: C(12, 2) - C(4, 2) = 60
A bit string contains 1's and 0's. How many different bit strings can be constructed given the restriction(s)? Length is 18. Has exactly two 1's in the first half. Has exactly four 1's in the second half.
For first half, choose two 1's from 9: C(9, 2)For second half, choose four 1's from 9: C(9, 4)Then, combine: C(9, 2) * C(9, 4)
How do you determine if a question asks about permutations or subset/combinations?
From gleaming from the question whether order is important or not. If it is, then the question is asking about permutations. If the order is not important, the question is asking about subset/combinations
How many different passwords are there that contain only digits and lower-case letters and satisfy the given restrictions? (a) Length is 6 and the password must contain at least one digit. (b) Length is 6 and the password must contain at least one digit and at least one letter.
Keyword "at least" means you will need to count by complement (a) The total number of passwords with no restrictions is 36⁶. The number of passwords with no digits (i.e., only letters) is 26⁶. Therefore the number of passwords with at least one digit is 36⁶ - 26⁶. (b) The total number of passwords with no restrictions is 36⁶. The number of passwords with no digits (i.e., only letters) is 26⁶. The number of passwords with no letters (i.e., only digits) is 10⁶. The number of passwords with no digits or no letters is 26⁶ + 10⁶. Therefore the number of passwords with at least one digit and at least one letter is 36⁶- 26⁶ - 10⁶
How many 8-bit strings have at least one 0?
Only one 8-bit string has no 0's: 11111111. The other 255 8-bit strings have at least one 0
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
P(12, 5) Suppose that the starters are placed into positions in the order in which they are selected. That is, the first player selected will play center, the second will play small forward, etc. The order in which the players are selected is important because the order determines which player plays which position. The number of ways to select an ordered list of five players from a set of 12 with no repetitions is P(12, 5).
The students in a class elect a president, vice president, secretary, and treasurer. There are 30 students in the class and no student can have more than one job. Note that it matters who is elected into which position. How many different outcomes are there from the election process?
P(30, 4) Select an order in which the class officers are elected: Pres, VP, Treas, Sec. The order in which the students are elected (in addition to which students are elected) is important because the order will determine which student gets which job. Therefore, a sequence of four kids are selected from the 30: (president, vice president, secretary, treasurer).
Dave will swim one day, run one day, and bike another day in a week. He does at most one activity on any particular day. How many ways are there for him to select his workout schedule (i.e. which activities he does which days)?
P(7, 3) First Dave picks his swimming day, then his running day, and then his biking day. Since each activity is different, the order in which he selects the days matters. Therefore the number of selections is P(7, 3).
How many ways are there to permute the letters in HAPPY? (Give a numerical answer)
String of length 5 with two P's, and one each of H, A, and Y:
How many ways are there to permute the letters in PEPPER? (Give a numerical answer)
String of length 6 with three P's, two E's, and one R:
How many 8-bits strings have exactly five 1's?
The five 1's can be put in any of the 8 locations throughout the string. Therefore you have 8 choose 5. Since the bottom half of the formula is r!(n-r)! if r is 5 then 5!(3!) would be the same if you had chosen r as 3....3!(5!).
Four people (John, Paul, George, and Ringo) are seated in a row on a bench. The number of ways to order the four people so that John is next to Paul is 12. How many ways are there to order the four people on the bench so that John is not next to Paul?
The number of ways to line up the 4 people so that John is not next to Paul is the total number of ways to line them up with no restrictions (4! = 24) minus the number of ways to line them up so that John is next to Paul (12). The answer is 24 - 12 = 12.
The math team has 6 girls and 4 boys. How many ways are there to select the two competitors if they are both girls?
The number of ways to select a set of 2 girls from 6 is C(6, 2) = 15
There are 10 kids on the math team. Two kids will be selected from the team to compete in the state competition. How many ways are there to select the 2 competitors?
The number of ways to select a set of 2 kids from 10 is C(10, 2) = 45
The math team has 6 girls and 4 boys. How many ways are there to select the two competitors so that at least one boy is chosen?
The number of ways to select the two competitors so that at least one boy is chosen is the total number of ways to select the competitors (without restrictions) [which is C(10, 2) = 45] minus the choices where nothing but girls are chosen which is C(6, 2) (=15). This guarantees that the remaining choices have at least one boy for the remaining number of ways to select the competitors. The answer is 45 - 15 = 30. **NOTE when you see "at least <one> <something> is chosen" think of a way to apply counting by complement.
Consider a function that maps 5-permutations from a set S = {1, 2, ..., 20} to 5-subsets from S. The function takes a 5-permutation and removes the ordering on the elements. How many 5-permutations map on to the subset {2, 5, 13, 14, 19}?
There are 5! ways to order five elements in a subset. Therefore, the function mapping 5-permutations to 5-subsets is k-to-1 for k = 5! = 120.
A camp offers 4 different activities for an elective: archery, hiking, crafts and swimming. The capacity in each activity is limited so that at most 35 kids can do archery, 20 can do hiking, 25 can do crafts and 20 can do swimming. There are 100 kids in the camp. How many ways are there to assign the kids to the activities?
There are C(100, 35) ways to select the kids who will do archery. Then there are C(65, 20) ways to select the kids who will go hiking. Then there are C(45, 25) ways to select the kids who will do crafts. The remaining 20 kids will go swimming. Putting together the selections by the product rule, the number of ways to assign kids to activities is:
Use the alternate way to solve the office assignment problem by applying the general product rule.
There are C(11, 3) ways to select the 3 people who will go into office A. After office A is filled, there are 8 people left from which to pick the 4 people for office B which results in C(8, 4) choices. After offices A and B are filled, there are four people left from which to pick the 4 people for office C which results in C(4, 4) choices.
120 pianists compete in a piano competition. (a) In the first round, 30 of the 120 are selected to go on to the next round. How many different outcomes are there for the first round? (b) In the second round, the judges select the first, second, third, fourth and fifth place winners of the competition from among the 30 pianists who advanced to the second round. How many outcomes are there for the second round of the competition?
There are C(120, 30) ways to select a subset of 30 pianists from a set of 120 pianists. The order in which the pianists are chosen does not matter, just the subset of 30 pianists selected. (b) Suppose that the pianists are selected in the order in which they rank. That is, the first pianist selected wins first place, the second pianist selected wins second place, etc. The order in which the pianists are selected is important because the order determines which pianists gets which place. The number of ways to select an ordered list of five pianists from a set of 30 with no repetitions is P(30, 5).
There are 20 members of a basketball team. The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
There are C(20, 12) ways to select a subset of 12 players from a set of 20 players. The order in which the players are chosen does not matter, just the subset of 12 players selected.
There are 30 boys and 35 girls that try out for a chorus. The choir director will select 10 girls and 10 boys from the children trying out. How many ways are there for the choir director to make his selection?
There are C(30, 10) ways to select a subset of 10 boys from a set of 30 boys. There are C(35, 10) ways to select a subset of 10 girls from a set of 35 girls. Since the choir director must select the girls and the boys for the chorus, the product rule applies, and there are a total of C(35, 10)*C(30, 10) ways to make the selection.
S = {a, b, c, d, e, f, g} How many subsets of S have either three or four elements?
There are C(7, 4) subsets with four elements and there are C(7, 3) subsets with three elements. A subset can not have three and four elements, so the sum rule applies. Therefore the number of subsets with three or four elements is C(7, 4) + C(7, 3).
How many ternary strings (digits 0,1, or 2) are there with exactly seven 0's, five 1's and four 2's?
There are a total of 7 (0's) + 5 (1's) + 4 (2's) = 16 characters in this string. The total number of ways to permute seven 0's, five 1's and four 2's is 16!/(7!5!4!).
S = {a, b, c, d, e, f, g} How many subsets of S have exactly four elements?
There are c(7, 4) ways to select a subset of 4 items from a set of 7 distinct items.
How many ways are there to permute the letters in the following word DISCRETE
There are eight letters in the word DISCRETE. The only repeated letters in DISCRETE are two E's. Therefore there are 8!/2! = 8!/2 = 4·7! ways to permute the letters in DISCRETE.
A family has four daughters. Their home has three bedrooms for the girls. Two of the bedrooms are only big enough for one girl. The other bedroom will have two girls. How many ways are there to assign the girls to bedrooms?
There are four ways to select the girl who will get the first single room. Then there are three ways to select the girl who will get the second single room. The two remaining girls are placed in the double room. Therefore the number of ways to assign girls to rooms is 4·3 = 12.
How many ways are there to permute the letters in the following word SUBSETS ?
There are seven letters in the word SUBSETS. The only repeated letters in SUBSETS are three S's. Therefore there are 7!/3! ways to permute the letters in SUBSETS.
How many ways are there to permute the letters in the following words NUMBER?
There are six letters in the word NUMBER. The letters in NUMBER are all distinct. Therefore there are 6! ways to permute the letters in NUMBER.
Let S = {a, b, c}. How many different 2-subsets from S are there?
Three The 2-subsets from S are {a, b}, {a, c}, {b, c}
What is counting by compliment?
a technique for counting the number of elements in a set S that have a property by counting the total number of elements in S and subtracting the number of elements in S that do not have the property. The principle of counting by complement can be written using set notation where P is the subset of elements in S that have the property.
What is a permutation with repetition?
an ordering of a set of items in which some of the items may be identical to each other To illustrate with a smaller example, there are 3! = 6 permutations of the letters CAT because the letters in CAT are all different. However, there are only 3 different ways to scramble the letters in DAD: ADD, DAD, DDA.
An equation is called an __________ if the equation holds for all values for which the expressions in the equation are well defined.
identity
