# Module 6: Basic Counting

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Module 6: Basic Counting Theme 1: Basic Counting Principle We start with two basic counting principles, namely, the sum rule and the multiplication rule. The Sum Rule: If there are n 1 different objects in the first set A 1, n objects in the second set A, :::, n m objects in the mth set A m, and if the sets A 1 ;A ;:::;A m are disjoint (i.e., A i A j = ; for any 1» i < j» m), then the total number of ways to select an object from one of the set is n 1 + n + + n m ; in other words, ja 1 [ A [ :::[ A m j = ja 1 j + ja j + + ja m j: The Multiplication Rule: Suppose a procedure can be broken into m successive (ordered) stages, with n 1 outcomes in the first stage, n outcomes in the second stage, :::, n m outcomes in the mth stage. If the number of outcomes at each stage is independent of the choices in previous stages, and if the composite outcomes are all distinct, then the total procedure has n 1 n n m different composite outcomes. A 1 ;:::;A m as follows Sometimes this rule can be phrased in terms of sets ja 1 A A m j = ja 1 j ja j ja m j: Example 1: Thereare40 students in an algebra class and 40 students in a geometry class. How many different students are in both classes combined? This problem is not well formulated and cannot be answered unless we are told how many students are taking both algebra and geometry. If there is not student taking both algebra and geometry, then by the sum rule the answer is But let us assume that there are 10 students taking both algebra and geometry. Then there are 30 students only in algebra, 30 students only in geometry, and 10 students in both algebra and geometry. Therefore, by the sum rule the total number of students is = 70. Example : There are boxes in a postal office labeled with an English letter (out of 6 English characters) and a positive integer not exceeding 80. How many boxes with different labels are possible? 1

2 The procedure of labeling boxes consists of two successive stages. In the first stage we assign 6 different English letters, and in the the second stage we assign 80 natural numbers (the second stage does not depend on the outcome of the first stage). Thus by the multiplication rule we have 6 80 = 080 different labels. Example 3: How many different bit strings are there of length five? We have here a procedure that assigns two values (i.e., zero or one) in five stages. Therefore, by the multiplication rule we have 5 =3different strings. Exercise 6A: How many binary strings of length 5 are there that start with a 1 and end with a 0? Example 4: Counting Functions. Let us consider functions from a set with m elements to a set with n elements. How many such functions are there? We can view this as a procedure of successive m stages with n outcomes in each stage, where the outcome of the next stage does not depend on the outcomes of the previous stages. By the multiplication rule there are n n n = n m functions. But, let us now count the number of one-to-one functions from a set of m elements to the set of n elements. Again, we deal here with a procedure of m successive stages. In the first stage we can assign n values. But in the second stage we can only assign n 1 values since for a one-toone function we are not allowed to select the value used before. In general, in the kth stage we have only n k +1elements at our disposal. Thus by (a generalized) multiplication rule we have n(n 1) (n m +1) one-to-one functions. Let us now consider some more sophisticated counting problems in which one must use a mixture of the sum and multiplication rules. Example 5: A valid file name must be six to eight characters long and each name must have at least one digit. How many file names can there be? If N is the total number of valid file names and N 6, N 7 and N 8 are, respectively, file names of length six, seven, and eight, then by the sum rule N = N 6 + N 7 + N 8 : Let us first estimate N 6. We compute it in an indirect way using the multiplication rule together with the sum rule. We first estimate the number of file names of length six without the constraint that there must be at least one digit. By the multiplication rule there are (6 + 10) 6 =36 6 file names. Now the number of file names that consists of only letters (no digits) is 6 6. We must subtract these since they are not allowed. Therefore (by the sum rule) In a similar way, we compute N 6 = = : N 7 = ;

3 N 8 = ; so that finally N = N 6 + N 7 + N 8 = : The next example illustrates the inclusion-exclusion principle that we already mentioned in Module : For two sets (not necessary disjoint) A and B the following holds ja [ Bj = jaj + jbj ja Bj (1) since in jaj + jbj the part ja Bj is counted twice, therefore we must subtract it. Example 6: How many bit strings of length eight start with 1 or end with two bits 00? We consider two tasks. The first one, constructing a string of length eight with the first bit equal to 1, can be done on 7 = 18 ways (by the multiplication rule after noticing that the first bit is set to be 1 and there are only seven free stages). In the second task we count the number of strings that end with 00. Again, by the multiplication rule there are 6 =3strings (since the last two stages are set to be 00). By adding these two numbers we would over count since both cases occur twice in this sum. To get it right, let us estimate the number of strings that starts with 1 and end with 00. Bythe multiplication rule we have 5 such strings (since three stages are set to be fixed). Therefore, by the inclusion-exclusion rule we find = 160; that is, it is the sum of strings with the first bit set to 1 and the the last two bits set to 00, minus the number of strings with the first first bit 1 and the last bits 00. 3

4 Theme : The Pigeonhole Principle Surprisingly many complex problems in combinatorics can be solved by an easy to state and prove principle called the pigeonhole principle. The Pigeonhole Principle. Ifk +1objects are placed into k boxes, then there is at least one box containing two or more of the objects. This principle is easy to prove by contradiction. Assume to the contrary that all boxes have at most one object. Since there are k boxes, we will end up with at most k objects, which contradicts the assumption stating that we have k +1objects. Example 7: Consider a set of 7 English words. There must be at least two words that begin with the same letter, since there are only 6 letters in the English alphabet. In some applications the following generalization of the pigeonhole principle is useful. Theorem 1 [Generalized Pigeonhole Principle] If N objects are placed into k boxes, then there is at least one box containing at least dn=ke objects, where dxe is the smallest integer larger or equal to x. Proof. Let us assume contrary that all boxes contain at most dn=ke 1 objects. Then the total number of objects is at most k (dn=ke 1) <k(n=k +1 1) = N which is impossible. Example 8: Consider a group of 100 students. Among them there are at least 9 who were born in the same month. Indeed, by the generalized pigeonhole principle with N = 100 and k =1we have at least d100=1e people born in the same month. Finally, we discuss two more sophisticated examples of the pigeonhole principle. Lemma 1. Among any n +1positive integers not exceeding n there must be an integer that divides one of the other integers from the set of n +1positive integers. Proof. Letthen +1integers be a 1 ;:::;a n+1. We represent every such an integer as a j = k j q j ; j =1; ;:::;n+1 where k j is a nonnegative integer and q j is an odd integer. For example, if a j =0, then we can write as 0 = 5, while a j =15= Certainly, the integers q 1 ;:::;q n+1 are odd integers smaller than n. Since there are only n odd integers smaller than n, it follows from the pigeonhole principle that two of the odd integers among n +1must be the same. Assume that q i = q j := q for i not equal to j. Then a i = k i q; a j = k j q: 4

5 Clearly, either a j divides a i or vice versa since two k i = k j = k i k j. The proof is completed. Exercise 6B: Justify that in any set of n +1positive integers not exceeding n there must be two that are relative prime (i.e., the greatest common divisor of both numbers is one). Example 9: Assume that in a group of six people, each pair of individuals consists of two friends or two enemies. We will show that there are either three mutual friends or three mutual enemies in the group. Indeed, let the group be labeled as A; B; C; D; E and F. Consider now the person labeled as A. The remaining five people can be grouped into friends or enemies of A. Of the five other people (other than A), there are either three or more who are friends of A, or three or more than are enemies of A. Indeed, when a set of 5 objects (persons) is divided into two groups (friends or enemies) there are at least d5=e =3elements in one of these groups. Consider first the group of friends of A. Call them B;C or D. If any of these three individuals are friends, then these two and A form the group of three mutual friends. Otherwise, B, C and D form a set of three mutual enemies. The proof in the case of three enemies of A proceeds in a similar manner. This last example is an instance of an important part of combinatorics called Ramsey theory. In general, Ramsey theory deals with the distribution of subsets of elements of sets. 5

6 Theme 3: Permutations and Combinations In computer science one often needs to know in how many ways one can arrange certain objects (e.g., how many inputs are there consisting of ten digits?). To answer these questions, we study here permutations and combinations the simplest arrangements of objects. A permutation of a set of distinct objects is an ordered arrangements of these objects. An ordered arrangements of r elements of a set is called an r-permutation. Example 10: LetS = fa; b; cg. Then abc, acb, bac, bca, cab, cba are permutations of S, while ab, ba, ac, ca, bc and cb are -permutations of S. It is not difficult to compute the number of r-permutations. Let P (n; r) be the number of r- permutations of a set with n distinct elements. Observe that we can choose the first element in the r-permutation in n ways, the second element in (n 1) (since after selecting the first element we can not use it again in the second choice), and so on, finally choosing the r-th element in n r +1ways. Therefore, by the multiplication rule the total number of r-permutations is P (n; r) =n(n 1) (n ) (n r +1)= r 1 Y i=0 (n i): () Above we use the product notation a 1 a a n = Q n i=1 a i introduced in Module. Example 11: On how many ways one can construct a three digits number with all different digits (e.g., 14 is a legitimate digit but 3 is not)? We recognize this problem as a 3-permutation, therefore the answer is = 70. In an r-permutation the order of elements is important (e.g., ab is different than ba), while in the r-combination is not. An r-combination of elements of a set is an unordered selection of r elements from the set (i.e., ab and ba are the same -combinations). The number of r-combinations of a set with n distinct elements is denoted by C(n; r) or Cn r. Thus the number of r-permutations is equal to the number of r-combinations times numbers of permutations (within each combinations), that is, P (n; r) =C(n; r) r! since every r-combination leads to r! =P (r;r) r-permutations. Example 1: Let S = fa; b; cg. Consider first -combinations. We have the following -combinations: that generate the following six -permutations fa; bg; fa; cg; fb; cg; (a; b); (b; a); (a; c); (c; a); (b; c); (c; b): 6

7 From the previous formula we immediately obtain C(n; r) = = = P (n; r) P (r;r) = n(n 1)(n ) (n r +1) r! n(n 1)(n ) (n r + 1)(n r)! n! r!(n r)! : r!(n r)! The first line above follows from the definition of r-permutations, while in the second line we multiply and divide by (n r)!, and finally in the third line we observed that n! =n(n 1)(n ) (n r+1)(n r)! = n(n 1)(n ) (n r+1) (n r) (n r 1) 1: In summary, we prove C(n; r) = n! r!(n r)! : (3) Exercise 6C: In how many ways one can create a four-letter word with all distinct letters (we assume there are 6 letters)? An astute reader should notice that C(n; r) was already introduced in Module 4 where we wrote it as: ψ! Cn r := C(n; n r) := : k Hereafter, we shall write C(n; r) for these numbers that are also called binomial coefficients or Newton s coefficients. In Module 4 we proved several properties of these coefficients algebraically. We now re-prove them using counting or combinatorial arguments. In particular, in Lemma of Module we proved algebraically that C(n; r) =C(n 1;r)+C(n 1;r 1): (4) We now re-establish it using counting (combinatorial) arguments. In order to obtain all r-combinations (= C(n; r)) we pick up one element from the set and put it aside. Call it z. Now we build all r- combinations from the set S fzg of size n 1. Clearly, we have C(n 1;r) such r-combinations. Let us now construct (r 1)-combinations from the set S fzg. We have C(n 1;r 1) such combinations. After adding the element z to such combinations we still have C(n 1;r 1) r- combinations, each different than in the first experiment (i.e., without using z). But combining these two r-combinations we obtain all possible r-combinations which is equal to C(n; r). We proved (4). In a similar fashion we can prove another identity established in Module 4, namely, C(n; r) =C(n; n r): 7

8 Indeed, there is one-to-one correspondence between r-combinations and (n r)-combinations: if fa;b;:::g is an r combination, then the corresponding (n r)-combination is S fa;b;:::g. In Module 4 we also proved that nx k=0 C(n; k) = n : We can re-establish it using counting arguments. Consider a set S of cardinality n. From Module we know that there are n subsets of S. The set of all subsets can be partitioned into subsets of size r, which are in fact r-combinations. There are C(n; r) combinations and they must sum to all subsets, which is n. Finally, we prove one new identity known as Vandermonde s Identity: C(m + n; r) = rx k=0 C(m; r k)c(n; k): (In words, the number of r-combinations among m + n elements is the sum of products of k combinations out of n and r k-combinations out of m.) We use a counting argument. Suppose that there are m items in one set and n items in another set. The total number of ways to select r items from the union of these sets is C(m + n; r). Another way of doing the same, is to select k items from the second set (we can do it in C(n; k) ways) and r k items from the first set (which can be done on C(m; r k) ways), where 0» k» r. By the multiplication rule these two actions can be done in C(m; r k)c(n; k) ways, hence the total number of ways to pick r elements is the sum over all k, and the Vandermonde identity is proved. 8

9 Theme 4: Generalized Permutations and Combinations In many counting problems, elements may be used repeatedly. For example, digits f0; 1;:::; 9g may be used more than once to form a valid number; letters can be repeatedly used in words (e.g., SUCCESS). In the previous section we assumed that the objects were distinguishable, while in this section we consider the case when some elements are indistinguishable. Finally, we also explain how to count the ways to place distinguishable elements in boxes (e.g., in how many ways poker hands can be dealt to four players). Permutations with Repetition The are n! permutations of n distinct (distinguishable) elements. But in how many ways we can obtain r-permutations when objects (elements) can be repeated? Example 13: Howmanywordsof7 characters can be created from 6 English letters? Observe that we do allow repetitions, so that SUCCESS is a valid word. By the multiplication rule we have 7 6 words but there are only words with all different letters. We can formulate the following general result. Consider r-permutations of a set with n elements when repetition is allowed. The number of r-permutations of such a set (with repetitions allowed) is n r : Indeed, we have r stages with n outcomes in each stage, hence by the multiplication rule the number of outcomes is n r. Combinations with Repetitions How many ways one can pick up (unordered) r elements from a set of n elements when repetitions are allowed? This is a harder problem, and we start with an example. Example 14: In a bag there are money bills of the following denominations: \$1; \$; \$5; \$10; \$0; \$50; and \$100: We are asked to select five bills. In how many ways we can do it assuming that the order in which the bills are chosen does not matter and there are at least five bills of each types? It is not C(7; 5) since we can pick up five bills of the same denomination. To solve this problem we apply an old combinatorial trick: We build an auxiliary device, that of a cash box with seven compartments, each one holding one type of bill. The bins containing the bills are separated by six dividers. Observe that selecting five bills corresponds to placing five markers (denoted usually as a star?) on the compartments holding the bills. For example, the following symbolic figure: 9

10 ????? corresponds to the case when one \$1 bill, three \$5 bills, and one \$0 bill are selected. Therefore, the number of ways to select five bills corresponds to the number of ways to arrange six bars (dividers) and five stars (markers). In other words, this amounts to selecting the position of the five stars from 11(= 6+5) positions. But this can be done in C(11; 5) = 11! 5!6! = 46 ways. This is the number of selecting five bills from a bag with seven types of bills. In general, let us select r-combinations from a set of n elements when repetition of elements is allowed. We represent this problem as a list of n 1 bars and r stars. These n 1 bars are used to mark n cells (bins). We assume that the ith cell contains a star whenever the ith element occurs in the combination. For instance, a 6-combination of a set of four elements has three bars and six stars. In particular,?????? corresponds to the combination containing exactly two of the first elements, none of the second element, one of the third element, and three of the fourth element of the set. In general, each different list containing n 1 bars and r stars corresponds to an r-combination of the set with n elements, when repetition is allowed. But the number of such lists is C(n 1+r;r) (5) which is also the number of r-combinations from the set of n elements when repetitions is allowed. Example 15: How many solutions does the following equation x 1 + x + x 3 + x 4 =15 have, where x 1 ;x ;x 3 and x 4 are nonnegative integers? Here is a solution to this problem. We assume we have four types labeled x 1 ;x ;x 3 and x 4. There are 15 items or units (since we are looking for an integer solution). Every time an item (unit) is selected it adds one to the type it picked it up. Observe that a solution corresponds to a way of selecting 15 items (units) from a set of four elements. Therefore, it is equal to 15-combinations with repetition allowed from a set with four elements. Thus by (5) we have C( ; 15) = C(18; 15) = C(18; 3) = solutions. (We recall that C(n; k) =C(n; n k).) =

11 Permutations of Sets with Indistinguishable Objects When counting some care must be exercised to avoid counting indistinguishable objects more than once. Example 16: How many different strings can be made by reordering the letters of the word TOT- TOS? If all letters in the word TOTTOS would be different, then the answer would be 5! but then we would over count. To avoid it, we observe that there are 6 positions. The letter T can be placed among these six positions in C(6; 3) times, while the letter O can be placed in the remaining positions in C(3; ) ways; finally S can be put in C(1; 1) ways. By the multiplication rule we have orderings, where we used the formula C(6; 3)C(3; )C(1; 1) = 6! = n! C(n; k) = k!(n k)! 3! 1! 3!3!!1! 1!0! 6! 3!!1! =60 learned before. We can obtain the same result in a different way. Observe that there are 6! permutations of six letters, however, there are 3! permutations in which permuting the letter T result in the same word; there! permutations of letter O that results in the same word. In summary, the number of different words is as before. 6! 3!!1! Let us now generalized the above example. Assume there are n objects with n 1 indistinguishable objects of type 1, n objects of type, :::, n k indistinguishable objects of type k. The number of different permutations are n! n 1!n! n : (6) k! There are many ways to prove this result. For example, we know that there n! permutations, but many of these permutations are the same since we have k classes of indistinguishable objects. How many permutations are the same due to n 1 indistinguishable objects. Obviously, there are n 1! such permutations of type 1, n! of type, :::, n k! of type k. Thus the result follows. Balls-and-Urns Model Finally, we consider throwing n distinguishable balls (objects) into k distinguishable urns (boxes). The combinatorial model will answer such questions as in how many ways five cards from a deck of 5 cards can be distributed to four players. 11

12 Consider the following example. There are n balls, and three boxes. We want to know in how many ways we can throw these n balls such that there are n 1 balls in the first box, n in the second box, and n 3 balls in the third box. Of course, there are C(n; n 1 ) of ways putting n 1 balls from a set of n balls into the first box. For every such an arrangement, the remaining n n 1 balls can be thrown in C(n n 1 ;n ) ways into the second box so that it contains n balls. Finally, the last box will have n 3 balls on C(n n 1 n ;n 3 ) ways. Therefore, by the multiplication rule we have C(n; n 1 )C(n n 1 ;n )C(n n 1 n ;n 3 ) = = n! (n n 1 )! (n n 1 n )! n 1!(n n 1 )! n!(n n 1 n )! n 3 (n n 1 n n 3 )! n! n 1!n!n 3! : In general, let n distinguishable objects be thrown into k distinguishable boxes with n i objects in the ith box, i =1; ;:::; k. Then, generalizing our example, we obtain n! n 1!n! n k!(n n 1 n ::: n k )! (7) ways to distribute these n objects among k boxes. Example 17: In how many ways we can distribute hands of 5 cards to each of four players from the deck of 5 cards? We may represent this problem as throwing 5 objects into four boxes each containing 5 cards. Thus the solution is 5! 5!5!5!5!3! : since every hand has 5 cards, and after the distribution of 4 5 cards there remain 3 cards. 1

13 Theme 5: Linear Recurrences 1 Consider the following problem, which was originally posed by Leonardo di Pisa, also known as Fibonacci, in the thirteenth century in his book Lieber abaci. A young pair of rabbits (one of each sex) is placed on an island. A pair of rabbits does not reproduce until they are months old. After they are month old, each pair of rabbits produces another pair each month. Denote F n the number of pairs of rabbits after n months, e.g. F 1 = F =1, F 3 =, F 4 =3, etc. (At the end of the first and second month there is only one pair, but at the end of the third, there is another one and at the end of the fourth once again, an additional one.) To find the number of pairs after n months we just have to add the number of rabbits in the previous month, F n 1, and the number of newborn pairs, which equals F n, since each newborn pair comes from a pair at least month old. Consequently, the sequence F n satisfies the recurrence relation F n = F n 1 + F n for n 3 together with the initial condition F 1 =1and F =. Of course, this recurrence relation and the initial condition uniquely determines the sequence F n. It should be mentioned that these numbers 1; 1; ; 3; 5; 8; 13; 1; 34; 55; 89;::: are called Fibonacci numbers. Sometimes the initial conditions for the Fibonacci numbers are F 0 =0 and F 1 =1(this has some theoretical advantages) but this causes just a shift of 1 in the index. Internet Exercise: Find on the internet two new applications of Fibonacci numbers and post a paragraph to the forum what you found out. The Fibonacci numbers occur in various counting problems. Example 18: Leta n denote the number of binary strings of length n with the property that there are no two subsequent ones: a 1 =: 0; 1 a =3: 00; 01; 10 a 3 =5: 000; 010; 001; 100; 101 : Again you can find a recurrence relation for a n. If the last bit of a sequence of length n (of that kind) is 0, then there are exactly a n 1 possible ways for the first n 1 bits. However, if the last bit is 1 then the (n 1)st bit has to be 0 and, hence, there are exactly a n possible ways for the first n 1 This material is more advanced and the student should take time to study it carefully. 13

14 letters. (For example, there are a 3 =5strings among which a =3ends with 0 and a 1 =with 1.) Consequently one obtains a n = a n 1 + a n for n 3. This is the same recurrence relation as for the Fibonacci numbers. condition is different: a 1 = F 3 and a = F 4. This implies a shift by two, that is, Only the initial a n = F n+ for all n 1. More generally we define: A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c a n + + c k a n k; where c 1 ;c ;:::;c k are real numbers, and c k 6=0. The recurrence relation is linear since the right-hand side is a sum of multiples of terms a n i. The recurrence relation is homogeneous since no terms occur that are not multiples of the a j s. The coefficients c j are all constant, rather than functions that depend on n. Thedegree is k because a n is expressed in terms of the previous k terms of the sequence. It is clear (by induction) that a sequence satisfying the recurrence relation in the definition is uniquely determined by its recurrence relation and the k initial conditions a 0 = C 0 ;a 1 = C 1 ;:::;a k 1 = C k 1: That is, once we set the first k 1 values, then the next values a n for n>kcan be computed from the recurrence. p Example 19: The recurrence relation a n = a n 1 is a linear homogeneous recurrence relation of degree one. The recurrence relation a n = a n 1 + a n is a linear homogeneous recurrence relation of degree two. The recurrence relation a n = a n 5 is a linear homogeneous recurrence relation of degree five. Example 0: The recurrence relation a n = a n 1 + a is not linear because there is term n a. n The recurrence relation a n =a n 1 +1is linear but not homogeneous because of the term 1 (which is not a multiply of a n ). The recurrence relation a n = na n 1 is linear but does not have constant coefficients. Exercise 6D: Is the following recurrence a n =6a n 1 +(n )a n +5 14

15 linear, homogeneous, with constant coefficients? The basic approach for solving linear homogeneous recurrence relations is to look for solutions of the form a n = r n ; where r is a constant. Note that a n = r n is a solution of the recurrence relation a n = c 1 a n 1 + c a n + c k a n k if and only if r n = c 1 r n 1 + c r n + + c k r n k : When both sides of this equation are divided by r n k and the right-hand side is subtracted from the left, we obtain the equivalent equation r k c 1 r k 1 c r k c k 1r c k =0: Consequently, the sequence a n = r n is a solution if and only if r is a solution of this last equation, which is called the characteristic equation of the recurrence relation. The solutions of this equation are called the characteristic roots of the recurrence relation. As we will see, these characteristic roots can be used to give an explicit formula for all solutions of the recurrence relation. We will consider in details the case of degree k =. The general case is similar but more involved and thus we will not stated. For k =the characteristic equation is just a quadratic equation of the form r c 1 r c =0. We recall that the quadratic equation r c 1 r c =0has two solutions p r c 1 ± c 1 +4c 1; = if c 1 +4c > 0; ifc 1 +4c =0, then the equation has one solution r 1 = c 1 =; otherwise there is no real solution to this equation. First, we consider the case when there are two distinct characteristic roots. Theorem. Let c 1 and c be real numbers. Suppose that r c 1 r c =0has two distinct roots r 1 and r. Then the sequence a n is a solution of the recurrence relation a n = c 1 a n 1 + c a n if and only if a n = ff 1 r n 1 + ff r n ; n 0; (8) where ff 1 and ff are constants. Proof. We must do two things to prove the theorem. First, it must be shown that if r 1 and r are the roots of the characteristic equation, and ff 1 and ff are constants, then the sequence a n = ff 1 r n 1 +ff r n is a solution of the recurrence relation. Second, it must be shown that if the sequence a n is a solution, then a n = ff 1 r n 1 + ff r n for some constants ff 1 and ff. 15

16 Now we will show that if a n = ff 1 r n 1 + ff r n, then the sequence a n is a solution of the recurrence equation. We have c 1 a n 1 + c a n = c 1 (ff 1 r n ff r n 1 )+c (ff 1 r n 1 + ff r n ) = ff 1 r n 1 (cr 1 + c )+ff r n (cr + c ) = ff 1 r n 1 + ff r n = a n : where the third lines is a consequence of the fact that since r 1 and r are roots of r c 1 r c =0, then r 1 = c 1r 1 + c and r = c 1r + c. This shows that the sequence a n with a n = ff 1 r n 1 + ff r n is a solution of the recurrence relation. To show that every solution a n of the recurrence relation a n = c 1 a n 1 + c a n is of the form a n = ff 1 r n 1 + ff r n for some constants ff 1 and ff, suppose that a n is a solution of the recurrence relation, and the initial conditions are a 0 = C 0 and a 1 = C 1. It will be shown that there are constants ff 1 and ff so that the sequence a n with a n = ff 1 r n 1 + ff r n satisfies the same initial conditions. This requires that a 0 = C 0 = ff 1 + ff ; a 1 = C 1 = ff 1 r 1 + ff r : We can solve these two equations for ff 1 and ff and get and ff 1 = C 1 C 0 r r 1 r ff C 0r 1 C 1 = ; r 1 r where these expressions depend on the fact that r 1 6= r.(when r 1 = r this theorem is not true.) Hence, with these values for ff 1 and ff, the sequence a n and the sequence ff 1 r n 1 + ff r n satisfy the two initial conditions a 0 = C 0 and a 1 = C 1. Since the recurrence relation and these initial conditions uniquely determine the sequence, it follows that a n = ff 1 r n 1 + ff r n. Example 1: We derive an explicit formula for the sequence of Fibonacci numbers F n defined by the recurrence relation F n = F n 1 + F n with initial conditions F 0 =0and F 1 =1. The characteristic equation of the Fibonacci recurrence is r r 1=0. Its solution is r 1 = 1+ p 1+4 = 1+ p 5 and r = 1 p 5 : Therefore, from (8) it follows that the Fibonacci numbers are given by ψ p! n ψ p 5 F n = ff 1 + ff! n 16

17 for some constants ff 1 and ff. The initial conditions F 0 = 0 and F 1 =1can be used to find these constants. We have F 0 = ff 1 + ff =0; ψ p 1+ 5 F 1 = ff 1! + ff ψ The solution of these simultaneous equations is given by 1 p 5! =1: ff 1 = 1 p 5 ff = p 1 : 5 Consequently, the Fibonacci numbers can be explicitly expressed as ψ p! n ψ p! n F n = p p : 5 5 Now we discuss the case when the characteristic equation has only one root (e.g., r r +1= (r 1) =0). We shall omit the proof. Theorem 3. Let c 1 and c real numbers. Suppose that r c 1 r c =0has only one root r 0.Then the sequence a n is a solution of the recurrence relation a n = c 1 a n 1 + c a n if and only if (for n 0), where ff 1 and ff are constants. a n = ff 1 r n 0 + ff nr n 0 Example : What is the solution of the following recurrence relation a n =6a n 1 9a n with initial conditions a 0 =1and a 1 =6? The only root of r 6r +9 = (r 3) = 0 is r = 3. Hence, the solution to this recurrence relation is a n = ff 1 3 n + ff n3 n for some constants ff 1 and ff. Using the initial conditions, it follows that a 0 =1=ff 1 ; a 1 =6=3ff 1 +3ff : Solving these two equations shows that ff 1 = 1 and ff = 1. Consequently, the solution to this recurrence relation is a n =3 n + n3 n =3 n (n +1): 17

### Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting

Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting Colin Stirling Informatics Slides originally by Kousha Etessami Colin Stirling (Informatics) Discrete Mathematics (Chapter 6) Today 1 /

### 1 Introduction. 2 Basic Principles. 2.1 Multiplication Rule. [Ch 9] Counting Methods. 400 lecture note #9

400 lecture note #9 [Ch 9] Counting Methods 1 Introduction In many discrete problems, we are confronted with the problem of counting. Here we develop tools which help us counting. Examples: o [9.1.2 (p.

### 94 Counting Solutions for Chapter 3. Section 3.2

94 Counting 3.11 Solutions for Chapter 3 Section 3.2 1. Consider lists made from the letters T, H, E, O, R, Y, with repetition allowed. (a How many length-4 lists are there? Answer: 6 6 6 6 = 1296. (b

### Discrete Mathematics: Homework 6 Due:

Discrete Mathematics: Homework 6 Due: 2011.05.20 1. (3%) How many bit strings are there of length six or less? We use the sum rule, adding the number of bit strings of each length to 6. If we include the

### Discrete Mathematics and its Applications Counting (2) Xiaocong ZHOU

Discrete Mathematics and its Applications Counting (2) Xiaocong ZHOU Department of Computer Science Sun Yat-sen University Feb. 2016 http://www.cs.sysu.edu.cn/ zxc isszxc@mail.sysu.edu.cn Xiaocong ZHOU

### Basics of Counting. The product rule. Product rule example. 22C:19, Chapter 6 Hantao Zhang. Sample question. Total is 18 * 325 = 5850

Basics of Counting 22C:19, Chapter 6 Hantao Zhang 1 The product rule Also called the multiplication rule If there are n 1 ways to do task 1, and n 2 ways to do task 2 Then there are n 1 n 2 ways to do

### It is not immediately obvious that this should even give an integer. Since 1 < 1 5

Math 163 - Introductory Seminar Lehigh University Spring 8 Notes on Fibonacci numbers, binomial coefficients and mathematical induction These are mostly notes from a previous class and thus include some

### Mathematical Foundations of Computer Science Lecture Outline

Mathematical Foundations of Computer Science Lecture Outline September 21, 2016 Example. How many 8-letter strings can be constructed by using the 26 letters of the alphabet if each string contains 3,4,

### Chapter 3. Distribution Problems. 3.1 The idea of a distribution. 3.1.1 The twenty-fold way

Chapter 3 Distribution Problems 3.1 The idea of a distribution Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls)

### NOTES ON COUNTING KARL PETERSEN

NOTES ON COUNTING KARL PETERSEN It is important to be able to count exactly the number of elements in any finite set. We will see many applications of counting as we proceed (number of Enigma plugboard

### MTH6109: Combinatorics. Professor R. A. Wilson

MTH6109: Combinatorics Professor R. A. Wilson Autumn Semester 2011 2 Lecture 1, 26/09/11 Combinatorics is the branch of mathematics that looks at combining objects according to certain rules (for example:

### Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

### 3. Recurrence Recursive Definitions. To construct a recursively defined function:

3. RECURRENCE 10 3. Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### Spring 2007 Math 510 Hints for practice problems

Spring 2007 Math 510 Hints for practice problems Section 1 Imagine a prison consisting of 4 cells arranged like the squares of an -chessboard There are doors between all adjacent cells A prisoner in one

### Notes on counting finite sets

Notes on counting finite sets Murray Eisenberg February 26, 2009 Contents 0 Introduction 2 1 What is a finite set? 2 2 Counting unions and cartesian products 4 2.1 Sum rules......................................

### Continued fractions and good approximations.

Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### Combinatorial Proofs

Combinatorial Proofs Two Counting Principles Some proofs concerning finite sets involve counting the number of elements of the sets, so we will look at the basics of counting. Addition Principle: If A

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### List of Standard Counting Problems

List of Standard Counting Problems Advanced Problem Solving This list is stolen from Daniel Marcus Combinatorics a Problem Oriented Approach (MAA, 998). This is not a mathematical standard list, just a

### CHAPTER 3 Numbers and Numeral Systems

CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,

### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### Section 6.4: Counting Subsets of a Set: Combinations

Section 6.4: Counting Subsets of a Set: Combinations In section 6.2, we learnt how to count the number of r-permutations from an n-element set (recall that an r-permutation is an ordered selection of r

### MA3059 Combinatorics and Graph Theory

MA3059 Combinatorics and Graph Theory Martin Stynes Department of Mathematics, UCC January 23, 2007 Abstract These notes are designed to accompany the textbook A first course in discrete mathematics by

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### A set is a Many that allows itself to be thought of as a One. (Georg Cantor)

Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### The Basics of Counting. Niloufar Shafiei

The Basics of Counting Niloufar Shafiei Counting applications Counting has many applications in computer science and mathematics. For example, Counting the number of operations used by an algorithm to

### Solution, Assignment #6, CSE 191 Fall, 2014

Solution, Assignment #6, CSE 191 Fall, 2014 General Guidelines: This assignment will NOT be collected, nor graded. However, you should carefully complete it as if it were to be graded. There will be a

### Problem Set 1 Solutions Math 109

Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twenty-four symmetries if reflections and products of reflections are allowed. Identify a symmetry which is

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

Chapter 6 Chapter Summary The Basics of Counting The Pigeonhole Principle Permutations and Combinations Binomial Coefficients and Identities Generalized Permutations and Combinations Generating Permutations

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### Elements of probability theory

2 Elements of probability theory Probability theory provides mathematical models for random phenomena, that is, phenomena which under repeated observations yield di erent outcomes that cannot be predicted

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS

1 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH INSTITUTE OF TECHNOLOGY CARLOW MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 Introduction Recurrence

### Chapter 7. Induction and Recursion. Part 1. Mathematical Induction

Chapter 7. Induction and Recursion Part 1. Mathematical Induction The principle of mathematical induction is this: to establish an infinite sequence of propositions P 1, P 2, P 3,..., P n,... (or, simply

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### Jan 17 Homework Solutions Math 151, Winter 2012. Chapter 2 Problems (pages 50-54)

Jan 17 Homework Solutions Math 11, Winter 01 Chapter Problems (pages 0- Problem In an experiment, a die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample

### Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.

Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to

### SYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me

SYSTEMS OF PYTHAGOREAN TRIPLES CHRISTOPHER TOBIN-CAMPBELL Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines

### Assigning Probabilities

What is a Probability? Probabilities are numbers between 0 and 1 that indicate the likelihood of an event. Generally, the statement that the probability of hitting a target- that is being fired at- is

### STRAND B: Number Theory. UNIT B2 Number Classification and Bases: Text * * * * * Contents. Section. B2.1 Number Classification. B2.

STRAND B: Number Theory B2 Number Classification and Bases Text Contents * * * * * Section B2. Number Classification B2.2 Binary Numbers B2.3 Adding and Subtracting Binary Numbers B2.4 Multiplying Binary

### Axiom A.1. Lines, planes and space are sets of points. Space contains all points.

73 Appendix A.1 Basic Notions We take the terms point, line, plane, and space as undefined. We also use the concept of a set and a subset, belongs to or is an element of a set. In a formal axiomatic approach

### 10. Graph Matrices Incidence Matrix

10 Graph Matrices Since a graph is completely determined by specifying either its adjacency structure or its incidence structure, these specifications provide far more efficient ways of representing a

### ON THE FIBONACCI NUMBERS

ON THE FIBONACCI NUMBERS Prepared by Kei Nakamura The Fibonacci numbers are terms of the sequence defined in a quite simple recursive fashion. However, despite its simplicity, they have some curious properties

### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### MATHEMATICAL THEORY FOR SOCIAL SCIENTISTS THE BINOMIAL THEOREM. (p + q) 0 =1,

THE BINOMIAL THEOREM Pascal s Triangle and the Binomial Expansion Consider the following binomial expansions: (p + q) 0 1, (p+q) 1 p+q, (p + q) p +pq + q, (p + q) 3 p 3 +3p q+3pq + q 3, (p + q) 4 p 4 +4p

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### 1 Mathematical Induction

Extra Credit Homework Problems Note: these problems are of varying difficulty, so you might want to assign different point values for the different problems. I have suggested the point values each problem

### Catalan Numbers. Thomas A. Dowling, Department of Mathematics, Ohio State Uni- versity.

7 Catalan Numbers Thomas A. Dowling, Department of Mathematics, Ohio State Uni- Author: versity. Prerequisites: The prerequisites for this chapter are recursive definitions, basic counting principles,

### Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent.

MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we

### Students in their first advanced mathematics classes are often surprised

CHAPTER 8 Proofs Involving Sets Students in their first advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most of the proofs they encounter are

### 1 Introduction to Counting

1 Introduction to Counting 1.1 Introduction In this chapter you will learn the fundamentals of enumerative combinatorics, the branch of mathematics concerned with counting. While enumeration problems can

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Discrete Mathematics: Homework 7 solution. Due: 2011.6.03

EE 2060 Discrete Mathematics spring 2011 Discrete Mathematics: Homework 7 solution Due: 2011.6.03 1. Let a n = 2 n + 5 3 n for n = 0, 1, 2,... (a) (2%) Find a 0, a 1, a 2, a 3 and a 4. (b) (2%) Show that

### A farmer raises rabbits. Each rabbit gives birth when 2 months old and then each month there after.

Fibonacci Problem: A farmer raises rabbits. Each rabbit gives birth when 2 months old and then each month there after. Question: How many rabbits will the farmer have after n months? Try it on small numbers.

### HMMT February Saturday 21 February Combinatorics

HMMT February 015 Saturday 1 February 015 1. Evan s analog clock displays the time 1:13; the number of seconds is not shown. After 10 seconds elapse, it is still 1:13. What is the expected number of seconds

### Lecture 3. Mathematical Induction

Lecture 3 Mathematical Induction Induction is a fundamental reasoning process in which general conclusion is based on particular cases It contrasts with deduction, the reasoning process in which conclusion

### Sets and Subsets. Countable and Uncountable

Sets and Subsets Countable and Uncountable Reading Appendix A Section A.6.8 Pages 788-792 BIG IDEAS Themes 1. There exist functions that cannot be computed in Java or any other computer language. 2. There

### 1. R In this and the next section we are going to study the properties of sequences of real numbers.

+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

### 3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.

Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible. 3.1.1 Definition A Dedekind

### Mathematical Induction

Mathematical Induction Victor Adamchik Fall of 2005 Lecture 2 (out of three) Plan 1. Strong Induction 2. Faulty Inductions 3. Induction and the Least Element Principal Strong Induction Fibonacci Numbers

### Matrix Algebra. Some Basic Matrix Laws. Before reading the text or the following notes glance at the following list of basic matrix algebra laws.

Matrix Algebra A. Doerr Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Some Basic Matrix Laws Assume the orders of the matrices are such that

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note 11

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note Conditional Probability A pharmaceutical company is marketing a new test for a certain medical condition. According

### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

### Mathematical Induction

Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: 1 + 3 + 5 +... + (2n 1) = n (2k 1). How might we proceed? The most natural

### Lecture 2 : Basics of Probability Theory

Lecture 2 : Basics of Probability Theory When an experiment is performed, the realization of the experiment is an outcome in the sample space. If the experiment is performed a number of times, different

### Notes on Determinant

ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 9-18/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without

### 9 abcd = dcba. 9 ( b + 10c) = c + 10b b + 90c = c + 10b b = 10c.

In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should

### Finite and Infinite Sets

Chapter 9 Finite and Infinite Sets 9. Finite Sets Preview Activity (Equivalent Sets, Part ). Let A and B be sets and let f be a function from A to B..f W A! B/. Carefully complete each of the following

### The Pigeonhole Principle

The Pigeonhole Principle 1 Pigeonhole Principle: Simple form Theorem 1.1. If n + 1 objects are put into n boxes, then at least one box contains two or more objects. Proof. Trivial. Example 1.1. Among 13

### Mathematics of Cryptography

CHAPTER 2 Mathematics of Cryptography Part I: Modular Arithmetic, Congruence, and Matrices Objectives This chapter is intended to prepare the reader for the next few chapters in cryptography. The chapter

### The set consisting of all natural numbers that are in A and are in B is the set f1; 3; 5g;

Chapter 5 Set Theory 5.1 Sets and Operations on Sets Preview Activity 1 (Set Operations) Before beginning this section, it would be a good idea to review sets and set notation, including the roster method

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### Discrete Structures for Computer Science: Counting, Recursion, and Probability

Discrete Structures for Computer Science: Counting, Recursion, and Probability Michiel Smid School of Computer Science Carleton University Ottawa, Ontario Canada michiel@scs.carleton.ca December 15, 2016

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

### Discrete mathematics

Discrete mathematics Petr Kovář petr.kovar@vsb.cz VŠB Technical University of Ostrava DiM 470-2301/01, Winter term 2015/2016 About this file This file is meant to be a guideline for the lecturer. Many

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

### Combinatorics: The Fine Art of Counting

Combinatorics: The Fine Art of Counting Lecture Notes Counting 101 Note to improve the readability of these lecture notes, we will assume that multiplication takes precedence over division, i.e. A / B*C

### Homework 3 (due Tuesday, October 13)

Homework (due Tuesday, October 1 Problem 1. Consider an experiment that consists of determining the type of job either blue-collar or white-collar and the political affiliation Republican, Democratic,

### Discrete Mathematics Lecture 5. Harper Langston New York University

Discrete Mathematics Lecture 5 Harper Langston New York University Empty Set S = {x R, x 2 = -1} X = {1, 3}, Y = {2, 4}, C = X Y (X and Y are disjoint) Empty set has no elements Empty set is a subset of

### Oh Yeah? Well, Prove It.

Oh Yeah? Well, Prove It. MT 43A - Abstract Algebra Fall 009 A large part of mathematics consists of building up a theoretical framework that allows us to solve problems. This theoretical framework is built

### Pigeonhole Principle Solutions

Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such

### Handout NUMBER THEORY

Handout of NUMBER THEORY by Kus Prihantoso Krisnawan MATHEMATICS DEPARTMENT FACULTY OF MATHEMATICS AND NATURAL SCIENCES YOGYAKARTA STATE UNIVERSITY 2012 Contents Contents i 1 Some Preliminary Considerations

### 8.7 Mathematical Induction

8.7. MATHEMATICAL INDUCTION 8-135 8.7 Mathematical Induction Objective Prove a statement by mathematical induction Many mathematical facts are established by first observing a pattern, then making a conjecture

### Week 9-10: Recurrence Relations and Generating Functions

Week 9-10: Recurrence Relations and Generating Functions March 31, 2015 1 Some number sequences An infinite sequence (or just a sequence for short is an ordered array a 0, a 1, a 2,..., a n,... of countably

### 1.3 Induction and Other Proof Techniques

4CHAPTER 1. INTRODUCTORY MATERIAL: SETS, FUNCTIONS AND MATHEMATICAL INDU 1.3 Induction and Other Proof Techniques The purpose of this section is to study the proof technique known as mathematical induction.

### Sets. A set is a collection of (mathematical) objects, with the collection treated as a single mathematical object.

Sets 1 Sets Informally: A set is a collection of (mathematical) objects, with the collection treated as a single mathematical object. Examples: real numbers, complex numbers, C integers, All students in

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express