CS计算机代考程序代写 AI Induction and recursion

Induction and recursion

Counting
Chapter 6
With Question/Answer Animations
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Chapter Summary
The Basics of Counting
The Pigeonhole Principle
Permutations and Combinations
Binomial Coefficients and Identities
Generalized Permutations and Combinations

The Basics of Counting
Section 6.1

Section Summary
The Product Rule
The Sum Rule
The Subtraction Rule
The Division Rule
Examples, Examples, and Examples
Tree Diagrams

Basic Counting Principles: The Product Rule
The Product Rule: A procedure can be broken down into a sequence of two tasks. There are n1 ways to do the first task and n2 ways to do the second task. Then there are n1∙n2 ways to do the procedure.

Example: How many bit strings of length seven are there?
Solution: ??

Example: How many bit strings of length seven are there?
Solution: Since each of the seven bits is either a 0 or a 1, the answer is 27 = 128.

The Product Rule
Example: How many different license plates can be made if each plate contains a sequence of three uppercase English letters followed by three digits?
Solution: ??

The Product Rule
Example: How many different license plates can be made if each plate contains a sequence of three uppercase English letters followed by three digits?
Solution: By the product rule,
there are 26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 = 17,576,000 different possible license plates.

Counting Functions
Counting Functions: How many functions are there from a set with m elements to a set with n elements?
Solution: Since a function represents a choice of one of the n elements of the codomain for each of the m elements in the domain, the product rule tells us that there are n ∙ n ∙ ∙ ∙ n = nm such functions.

Counting One-to-One Functions: How many one-to-one functions are there from a set with m elements to one with n elements?
Solution: ??

Counting One-to-One Functions: How many one-to-one functions are there from a set with m elements to one with n elements?
Solution: Suppose the elements in the domain are a1, a2,…, am. There are n ways to choose the value of a1 and n−1 ways to choose a2, etc. The product rule tells us that there are n(n−1) (n−2)∙∙∙(n−m +1) such functions.

Telephone Numbering Plan
Example: The North American numbering plan (NANP) specifies that a telephone number consists of 10 digits, consisting of a three-digit area code, a three-digit office code, and a four-digit station code. There are some restrictions on the digits.
Let X denote a digit from 0 through 9.
Let N denote a digit from 2 through 9.
Let Y denote a digit that is 0 or 1.
In the old plan (in use in the 1960s) the format was NYX-NNX-XXX.
In the new plan, the format is NXX-NXX-XXX.
How many different telephone numbers are possible under the old plan and the new plan?

Solution: Use the Product Rule.
There are 8 ∙2 ∙10 = 160 area codes with the format NYX.
There are 8 ∙10 ∙10 = 800 area codes with the format NXX.
There are 8 ∙8 ∙10 = 640 office codes with the format NNX.
There are 10 ∙10 ∙10 ∙10 = 10,000 station codes with the format XXXX.
Number of old plan telephone numbers: 160 ∙640 ∙10,000 = 1,024,000,000.
Number of new plan telephone numbers: 800 ∙800 ∙10,000 = 6,400,000,000.

Counting Subsets of a Finite Set
Counting Subsets of a Finite Set: Use the product rule to show that the number of different subsets of a finite set S is 2|S|. (In Section 5.1, mathematical induction was used to prove this same result.)
Solution: When the elements of S are listed in an arbitrary order, there is a one-to-one correspondence between subsets of S and bit strings of length |S|. When the ith element is in the subset, the bit string has a 1 in the ith position and a 0 otherwise.

By the product rule, there are 2|S| such bit strings, and therefore 2|S| subsets.

Product Rule in Terms of Sets
If A1, A2, … , Am are finite sets, then the number of elements in the Cartesian product of these sets is the product of the number of elements of each set.
The task of choosing an element in the Cartesian product A1 ⨉ A2 ⨉ ∙∙∙ ⨉ Am is done by choosing an element in A1, an element in A2 , …, and an element in Am.
By the product rule, it follows that:

|A1 ⨉ A2 ⨉ ∙∙∙ ⨉ Am |= |A1| ∙ |A2| ∙ ∙∙∙ ∙ |Am|.

Basic Counting Principles: The Sum Rule
The Sum Rule: If a task can be done either in one of n1 ways or in one of n2, where none of the set of n1 ways is the same as any of the n2 ways, then there are n1 + n2 ways to do the task.
Example: The mathematics department must choose either a student or a faculty member as a representative for a university committee. How many choices are there for this representative if there are 37 members of the mathematics faculty and 83 mathematics majors and no one is both a faculty member and a student.
Solution: ??

The Sum Rule in terms of sets.
The sum rule can be phrased in terms of sets.
|A ∪ B|= |A| + |B| as long as A and B are disjoint sets.
Or more generally,

The case where the sets have elements in common will be discussed when we consider the subtraction rule and taken up fully in Chapter 8.

|A1 ∪ A2 ∪ ∙∙∙ ∪ Am |= |A1| + |A2| + ∙∙∙ + |Am|
when Ai ∩ Aj = ∅ for all i, j.

Combining the Sum and Product Rule
Example: Suppose statement labels in a programming language can be either a single letter or a letter followed by a digit. Find the number of possible labels.
Solution: ??

Counting Passwords
Combining the sum and product rule allows us to solve more complex problems.
Example: Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there?

Solution: ??

Solution: Let P be the total number of passwords, and let P6, P7, and P8 be the passwords of length 6, 7, and 8.
By the sum rule P = P6 + P7 +P8.
To find each of P6, P7, and P8 , we find the number of passwords of the specified length composed of letters and digits and subtract the number composed only of letters. We find that:

P6 = 366 − 266 =2,176,782,336 − 308,915,776 =1,867,866,560.
P7 = 367 − 267 =
78,364,164,096 − 8,031,810,176 = 70,332,353,920.
P8 = 368 − 268 =
2,821,109,907,456 − 208,827,064,576 =2,612,282,842,880.

Consequently, P = P6 + P7 +P8 = 2,684,483,063,360.

Internet Addresses
Version 4 of the Internet Protocol (IPv4) uses 32 bits.

Class A Addresses: used for the largest networks, a 0,followed by a 7-bit netid and a 24-bit hostid.
Class B Addresses: used for the medium-sized networks, a 10,followed by a 14-bit netid and a 16-bit hostid.
Class C Addresses: used for the smallest networks, a 110,followed by a 21-bit netid and a 8-bit hostid.
Neither Class D nor Class E addresses are assigned as the address of a computer on the internet. Only Classes A, B, and C are available.
1111111 is not available as the netid of a Class A network.
Hostids consisting of all 0s and all 1s are not available in any network.

Counting Internet Addresses
Example: How many different IPv4 addresses are available for computers on the internet?
Solution: Use both the sum and the product rule. Let x be the number of available addresses, and let xA, xB, and xC denote the number of addresses for the respective classes.
To find, xA: 27 − 1 = 127 netids. 224 − 2 = 16,777,214 hostids.
xA = 127∙ 16,777,214 = 2,130,706,178.
To find, xB: 214 = 16,384 netids. 216 − 2 = 16,534 hostids.
xB = 16,384 ∙ 16, 534 = 1,073,709,056.
To find, xC: 221 = 2,097,152 netids. 28 − 2 = 254 hostids.
xC = 2,097,152 ∙ 254 = 532,676,608.
Hence, the total number of available IPv4 addresses is
x = xA + xB + xC
= 2,130,706,178 + 1,073,709,056 + 532,676,608
= 3, 737,091,842.
Not Enough Today !!
The newer IPv6 protocol solves the problem of too few addresses.

Basic Counting Principles: Subtraction Rule
Subtraction Rule: If a task can be done either in one of n1 ways or in one of n2 ways, then the total number of ways to do the task is n1 + n2 minus the number of ways to do the task that are common to the two different ways.
Also known as, the principle of inclusion-exclusion:

Counting Bit Strings
Example: How many bit strings of length eight either start with a 1 bit or end with the two bits 00?
Solution:??

Counting Bit Strings
Example: How many bit strings of length eight either start with a 1 bit or end with the two bits 00?
Solution: Use the subtraction rule.
Number of bit strings of length eight that start with a 1 bit: 27 = 128
Number of bit strings of length eight that end with bits 00: 26 = 64
Number of bit strings of length eight that start with a 1 bit and end with bits 00 : 25 = 32
Hence, the number is 128 + 64 − 32 = 160.

Tree Diagrams
Tree Diagrams: We can solve many counting problems through the use of tree diagrams, where a branch represents a possible choice and the leaves represent possible outcomes.
Example: Suppose that “I Love Discrete Math” T-shirts come in five different sizes: S,M,L,XL, and XXL. Each size comes in four colors (white, red, green, and black), except XL, which comes only in red, green, and black, and XXL, which comes only in green and black. What is the minimum number of shirts that the campus book store needs to stock to have one of each size and color available?
Solution: Draw the tree diagram.

The store must stock 17 T-shirts.

The Pigeonhole Principle
Section 6.2

Section Summary
The Pigeonhole Principle
The Generalized Pigeonhole Principle

The Pigeonhole Principle
If a flock of 20 pigeons roosts in a set of 19 pigeonholes, one of the pigeonholes must have more than 1 pigeon.

Pigeonhole Principle: If k is a positive integer and k + 1 objects are placed into k boxes, then at least one box contains two or more objects.
Proof: We use a proof by contraposition. Suppose none of the k boxes has more than one object. Then the total number of objects would be at most k. This contradicts the statement that we have k + 1 objects.

The Pigeonhole Principle
Corollary 1: A function f from a set with k + 1 elements to a set with k elements is not one-to-one.
Proof: Use the pigeonhole principle.
Create a box for each element y in the codomain of f .
Put in the box for y all of the elements x from the domain such that f(x) = y.
Because there are k + 1 elements and only k boxes, at least one box has two or more elements.
Hence, f can’t be one-to-one.

Pigeonhole Principle
Example: Among any group of 367 people, there must be at least two with the same birthday, because there are only 366 possible birthdays.

Example (optional): Show that for every integer n there is a multiple of n that has only 0s and 1s in its decimal expansion.
Solution: Let n be a positive integer. Consider the n + 1 integers 1, 11, 111, …., 11…1 (where the last has n + 1 1s). There are n possible remainders when an integer is divided by n. By the pigeonhole principle, when each of the n + 1 integers is divided by n, at least two must have the same remainder. Subtract the smaller from the larger and the result is a multiple of n that has only 0s and 1s in its decimal expansion.

The Generalized Pigeonhole Principle
The Generalized Pigeonhole Principle: If N objects are placed into k boxes, then there is at least one box containing at least ⌈N/k⌉ objects.
Proof: We use a proof by contraposition. Suppose that none of the boxes contains more than ⌈N/k⌉ − 1 objects. Then the total number of objects is at most

where the inequality ⌈N/k⌉ < ⌈N/k⌉ + 1 has been used. This is a contradiction because there are a total of n objects. Example: Among 100 people there are at least ⌈100/12⌉ = 9 who were born in the same month. The Generalized Pigeonhole Principle Example: a) How many cards must be selected from a standard deck of 52 cards to guarantee that at least three cards of the same suit are chosen? b) How many must be selected to guarantee that at least three hearts are selected? Solution: ?? The Generalized Pigeonhole Principle Example: a) How many cards must be selected from a standard deck of 52 cards to guarantee that at least three cards of the same suit are chosen? b) How many must be selected to guarantee that at least three hearts are selected? Solution: a) We assume four boxes; one for each suit. Using the generalized pigeonhole principle, at least one box contains at least ⌈N/4⌉ cards. At least three cards of one suit are selected if ⌈N/4⌉ ≥3. The smallest integer N such that ⌈N/4⌉ ≥3 is N = 2 ∙ 4 + 1 = 9. b) A deck contains 13 hearts and 39 cards which are not hearts. So, if we select 41 cards, we may have 39 cards which are not hearts along with 2 hearts. However, when we select 42 cards, we must have at least three hearts. (Note that the generalized pigeonhole principle is not used here.) Permutations and Combinations Section 6.3 Section Summary Permutations Combinations Combinatorial Proofs Permutations Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement of r elements of a set is called an r-permuation. Example: Let S = {1,2,3}. The ordered arrangement 3,1,2 is a permutation of S. The ordered arrangement 3,2 is a 2-permutation of S. The number of r-permuatations of a set with n elements is denoted by P(n,r). The 2-permutations of S = {1,2,3} are 1,2; 1,3; 2,1; 2,3; 3,1; and 3,2. Hence, P(3,2) = 6. A Formula for the Number of Permutations Theorem 1: If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are P(n, r) = n(n − 1)(n − 2) ∙∙∙ (n − r + 1) r-permutations of a set with n distinct elements. Proof: Use the product rule. The first element can be chosen in n ways. The second in n − 1 ways, and so on until there are (n − ( r − 1)) ways to choose the last element. Note that P(n,0) = 1, since there is only one way to order zero elements. Corollary 1: If n and r are integers with 1 ≤ r ≤ n, then Solving Counting Problems by Counting Permutations Example: How many ways are there to select a first-prize winner, a second prize winner, and a third-prize winner from 100 different people who have entered a contest? Solution: ?? Solving Counting Problems by Counting Permutations Example: How many ways are there to select a first-prize winner, a second prize winner, and a third-prize winner from 100 different people who have entered a contest? Solution: P(100,3) = 100 ∙ 99 ∙ 98 = 970,200 Solving Counting Problems by Counting Permutations (continued) Example: Suppose that a saleswoman has to visit eight different cities. She must begin her trip in a specified city, but she can visit the other seven cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities? Solution: ?? Solving Counting Problems by Counting Permutations (continued) Example: Suppose that a saleswoman has to visit eight different cities. She must begin her trip in a specified city, but she can visit the other seven cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities? Solution: The first city is chosen, and the rest are ordered arbitrarily. Hence the orders are: 7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040 If she wants to find the tour with the shortest path that visits all the cities, she must consider 5040 paths! Solving Counting Problems by Counting Permutations (continued) Example: How many permutations of the letters ABCDEFGH contain the string ABC ? Solution:?? Solving Counting Problems by Counting Permutations (continued) Example: How many permutations of the letters ABCDEFGH contain the string ABC ? Solution: We solve this problem by counting the permutations of six objects, ABC, D, E, F, G, and H. 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 Combinations Definition: An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements. The number of r-combinations of a set with n distinct elements is denoted by C(n, r). The notation is also used and is called a binomial coefficient. (We will see the notation again in the binomial theorem in Section 6.4.) Example: Let S be the set {a, b, c, d}. Then {a, c, d} is a 3-combination from S. It is the same as {d, c, a} since the order listed does not matter. C(4,2) = 6 because the 2-combinations of {a, b, c, d} are the six subsets {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}. Combinations Theorem 2: The number of r-combinations of a set with n elements, where n ≥ r ≥ 0, equals Proof: By the product rule P(n, r) = C(n,r) ∙ P(r,r). Therefore, Combinations Example: How many poker hands of five cards can be dealt from a standard deck of 52 cards? Also, how many ways are there to select 47 cards from a deck of 52 cards? Solution: ?? This is a special case of a general result. → Combinations Example: How many poker hands of five cards can be dealt from a standard deck of 52 cards? Also, how many ways are there to select 47 cards from a deck of 52 cards? Solution: Since the order in which the cards are dealt does not matter, the number of five card hands is: The different ways to select 47 cards from 52 is This is a special case of a general result. → Combinations Corollary 2: Let n and r be nonnegative integers with r ≤ n. Then C(n, r) = C(n, n − r). Proof: From Theorem 2, it follows that and Hence, C(n, r) = C(n, n − r). This result can be proved without using algebraic manipulation. → Combinations Example: How many ways are there to select five players from a 10-member tennis team to make a trip to a match at another school. Solution: By Theorem 2, the number of combinations is Example: A group of 30 people have been trained as astronauts to go on the first mission to Mars. How many ways are there to select a crew of six people to go on this mission? Solution: By Theorem 2, the number of possible crews is Binomial Coefficients and Identities Section 6.4 Section Summary The Binomial Theorem Pascal’s Identity and Triangle Powers of Binomial Expressions Definition: A binomial expression is the sum of two terms, such as x + y. (More generally, these terms can be products of constants and variables.) We can use counting principles to find the coefficients in the expansion of (x + y)n where n is a positive integer. To illustrate this idea, we first look at the process of expanding (x + y)3. (x + y) (x + y) (x + y) expands into a sum of terms that are the product of a term from each of the three sums. Terms of the form x3, x2y, x y2, y3 arise. The question is what are the coefficients? To obtain x3 , an x must be chosen from each of the sums. There is only one way to do this. So, the coefficient of x3 is 1. To obtain x2y, an x must be chosen from two of the sums and a y from the other. There are ways to do this and so the coefficient of x2y is 3. To obtain xy2, an x must be chosen from of the sums and a y from the other two . There are ways to do this and so the coefficient of xy2 is 3. To obtain y3 , a y must be chosen from each of the sums. There is only one way to do this. So, the coefficient of y3 is 1. We have used a counting argument to show that (x + y)3 = x3 + 3x2y + 3x y2 + y3 . Next we present the binomial theorem gives the coefficients of the terms in the expansion of (x + y)n . Binomial Theorem Binomial Theorem: Let x and y be variables, and n a nonnegative integer. Then: Proof: We use combinatorial reasoning . The terms in the expansion of (x + y)n are of the form xn−jyj for j = 0,1,2,…,n. To form the term xn−jyj, it is necessary to choose n−j xs from the n sums. Therefore, the coefficient of xn−jyj is which equals . Using the Binomial Theorem Example: What is the coefficient of x12y13 in the expansion of (2x − 3y)25? Solution: ??? Using the Binomial Theorem Example: What is the coefficient of x12y13 in the expansion of (2x − 3y)25? Solution: We view the expression as (2x +(−3y))25. By the binomial theorem Consequently, the coefficient of x12y13 in the expansion is obtained when j = 13. A Useful Identity Corollary 1: With n ≥0, Proof (using binomial theorem): With x = 1 and y = 1, from the binomial theorem we see that: Proof (combinatorial): Consider the subsets of a set with n elements. There are subsets with zero elements, with one element, with two elements, …, and with n elements. Therefore the total is Since, we know that a set with n elements has 2n subsets, we conclude: Pascal’s Identity Pascal’s Identity: If n and k are integers with n ≥ k ≥ 0, then Proof (combinatorial): Let T be a set where |T| = n + 1, a ∊T, and S = T − {a}. There are subsets of T containing k elements. Each of these subsets either: contains a with k − 1 other elements, or contains k elements of S and not a. There are subsets of k elements that contain a, since there are subsets of k − 1 elements of S, subsets of k elements of T that do not contain a, because there are subsets of k elements of S. Hence, Blaise Pascal (1623-1662) See Exercise 19 for an algebraic proof. Pascal’s Triangle The nth row in the triangle consists of the binomial coefficients , k = 0,1,….,n. By Pascal’s identity, adding two adjacent bionomial coefficients results is the binomial coefficient in the next row between these two coefficients. Generalized Permutations and Combinations Section 6.5 Section Summary Permutations with Repetition Combinations with Repetition Permutations with Indistinguishable Objects Distributing Objects into Boxes Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Proof: There are n ways to select an element of the set for each of the r positions in the r-permutation when repetition is allowed. Hence, by the product rule there are nr r-permutations with repetition. Example: How many strings of length r can be formed from the uppercase letters of the English alphabet? Solution: ?? Permutations with Repetition Theorem 1: The number of r-permutations of a set of n objects with repetition allowed is nr. Proof: There are n ways to select an element of the set for each of the r positions in the r-permutation when repetition is allowed. Hence, by the product rule there are nr r-permutations with repetition. Example: How many strings of length r can be formed from the uppercase letters of the English alphabet? Solution: The number of such strings is 26r, which is the number of r-permutations of a set with 26 elements. Combinations with Repetition Example: How many ways are there to select five bills from a box containing at least five of each of the following denominations: $1, $2, $5, $10, $20, $50, and $100? Solution: Place the selected bills in the appropriate position of a cash box illustrated below: continued → Combinations with Repetition Some possible ways of placing the five bills: The number of ways to select five bills corresponds to the number of ways to arrange six bars and five stars in a row. This is the number of unordered selections of 5 objects from a set of 11. Hence, there are ways to choose five bills with seven types of bills. Combinations with Repetition Theorem 2: The number 0f r-combinations from a set with n elements when repetition of elements is allowed is C(n + r – 1,r) = C(n + r – 1, n –1). Proof: Each r-combination of a set with n elements with repetition allowed can be represented by a list of n –1 bars and r stars. The bars mark the n cells containing a star for each time the ith element of the set occurs in the combination. The number of such lists is C(n + r – 1, r), because each list is a choice of the r positions to place the stars, from the total of n + r – 1 positions to place the stars and the bars. This is also equal to C(n + r – 1, n –1), which is the number of ways to place the n –1 bars. Combinations with Repetition Example: How many solutions does the equation x1 + x2 + x3 = 11 have, where x1 , x2 and x3 are nonnegative integers? Solution:?? Combinations with Repetition Example: How many solutions does the equation x1 + x2 + x3 = 11 have, where x1 , x2 and x3 are nonnegative integers? Solution: Each solution corresponds to a way to select 11 items from a set with three elements; x1 elements of type one, x2 of type two, and x3 of type three. By Theorem 2 it follows that there are solutions. Combinations with Repetition Example: Suppose that a cookie shop has four different kinds of cookies. How many different ways can six cookies be chosen? Solution:?? Combinations with Repetition Example: Suppose that a cookie shop has four different kinds of cookies. How many different ways can six cookies be chosen? Solution: The number of ways to choose six cookies is the number of 6-combinations of a set with four elements. By Theorem 2 is the number of ways to choose six cookies from the four kinds. Summarizing the Formulas for Counting Permutations and Combinations with and without Repetition Permutations with Indistinguishable Objects Example: How many different strings can be made by reordering the letters of the word SUCCESS. Solution: ?? Permutations with Indistinguishable Objects Example: How many different strings can be made by reordering the letters of the word SUCCESS. Solution: There are seven possible positions for the three Ss, two Cs, one U, and one E. The three Ss can be placed in C(7,3) different ways, leaving four positions free. The two Cs can be placed in C(4,2) different ways, leaving two positions free. The U can be placed in C(2,1) different ways, leaving one position free. The E can be placed in C(1,1) way. By the product rule, the number of different strings is: The reasoning can be generalized to the following theorem. → Permutations with Indistinguishable Objects Theorem 3: The number of different permutations of n objects, where there are n1 indistinguishable objects of type 1, n2 indistinguishable objects of type 2, …., and nk indistinguishable objects of type k, is: Proof: By the product rule the total number of permutations is: C(n, n1 ) C(n − n1, n2 ) ∙∙∙ C(n − n1 − n2 − ∙∙∙ − nk, nk) since: The n1 objects of type one can be placed in the n positions in C(n, n1 ) ways, leaving n − n1 positions. Then the n2 objects of type two can be placed in the n − n1 positions in C(n − n1, n2 ) ways, leaving n − n1 − n2 positions. Continue in this fashion, until nk objects of type k are placed in C(n − n1 − n2 − ∙∙∙ − nk, nk) ways. The product can be manipulated into the desired result as follows: Distributing Objects into Boxes Many counting problems can be solved by counting the ways objects can be placed in boxes. The objects may be either different from each other (distinguishable) or identical (indistinguishable). The boxes may be labeled (distinguishable) or unlabeled (indistinguishable). Distributing Objects into Boxes Distinguishable objects and distinguishable boxes. There are n!/(n1!n2! ∙∙∙nk!) ways to distribute n distinguishable objects into k distinguishable boxes. (See Exercises 47 and 48 for two different proofs.) Example: There are 52!/(5!5!5!5!32!) ways to distribute hands of 5 cards each to four players. Indistinguishable objects and distinguishable boxes. There are C(n + r − 1, n − 1) ways to place r indistinguishable objects into n distinguishable boxes. Proof based on one-to-one correspondence between n-combinations from a set with k-elements when repetition is allowed and the ways to place n indistinguishable objects into k distinguishable boxes. Example: There are C(8 + 10 − 1, 10) = C(17,10) = 19,448 ways to place 10 indistinguishable objects into 8 distinguishable boxes. Distributing Objects into Boxes Distinguishable objects and indistinguishable boxes. Example: There are 14 ways to put four employees into three indistinguishable offices (see Example 10). There is no simple closed formula for the number of ways to distribute n distinguishable objects into j indistinguishable boxes. See the text for a formula involving Stirling numbers of the second kind. Indistinguishable objects and indistinguishable boxes. Example: There are 9 ways to pack six copies of the same book into four identical boxes (see Example 11). The number of ways of distributing n indistinguishable objects into k indistinguishable boxes equals pk(n), the number of ways to write n as the sum of at most k positive integers in increasing order. No simple closed formula exists for this number. /docProps/thumbnail.jpeg

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