Elements of Combinatorics (Continuation) 1. Pigeonhole Principle Theorem. If pigeons are placed into pigeonholes and there are more pigeons than pigeonholes,

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Presentation transcript:

Elements of Combinatorics (Continuation) 1

Pigeonhole Principle Theorem. If pigeons are placed into pigeonholes and there are more pigeons than pigeonholes, then some pigeonhole must contain at least two pigeons. In general, if the number of pigeons is more than k times the number of pigeonholes, then some pigeonhole must contain at least k+1 pigeons. 2

Pigeonhole Principle 3

Example How many people must be selected from a collection of 15 married couples to ensure that at least two of the persons chosen are married to each other? Answer: 16 4

Multiplication Principle Theorem. Consider a procedure that is composed of a sequence of k steps. Suppose that the first step can be performed in n 1 ways; and for each of these, the second step can be performed in n 2 ways; and in general, no matter how the preceding steps are performed, the ith step can be performed in n i ways ( i =2,3,…,k). Then the number of different ways in which the entire procedure can be performed is 5

Example 1 Suppose a student has to choose two elective classes: 1 class out of 3 in the field of study area and 1 class out of 2 from the core curriculum. How many possible combinations of the chosen classes exist? Answer: according to the multiplication principle there are 6=3x2 combinations. 6

Example 2 There are 3 professors. They have to teach 6 classes. Each of them has to teach exactly 2 classes. How many possibilities exist for professors’ assignment? Solution: The first professor may choose 2 classes from 6 ( ), the second one may choose 2 from 4 remaining classes ( ). The third professor has no choice and must teach 2 remaining classes. According to the multiplication principle there are possibilities for professors’ assignment. 7

Addition Principle Theorem. Suppose that there are k sets of elements with n 1 elements in the first set, n 2 elements in the second set, etc. If all of the elements are distinct (that is, if all pairs of the k sets are disjoint), then the number of elements in the union of the sets is 8

Example Suppose a student has to choose one elective class. It may be either a class in the field of study (out of 6 offered) or a class from the core curriculum (out of 4 offered). How many different classes may the student choose? Answer: according to the addition principle there are 10=6+4 classes. 9

PERMUTATIONS, COMBINATIONS, AND ARRANGEMENTS WITH REPETITIONS 10

Permutations with Repetitions Theorem. Let S be a collection containing n objects of k different types (objects of the same type are indistinguishable, and objects of different types are distinguishable). Suppose that each object is of exactly one type and that there are n 1 objects of type 1, n 2 objects of type 2, and, in general, n i objects of type i. Then the number of different permutations of the objects in S is 11

Example Suppose there is a capstone class with 9 students enrolled. The instructor has decided to give students 3 projects and to divide them in 3 groups, respectively. Since the projects require different work efforts and resources, the 1 st group has to contain 3 students, the 2 nd one – 4 students, and the 3 rd one – 2 students. How many different opportunities there exist to form the groups? Solution: according to the permutations with repetitions theorem there are opportunities. 12

Arrangements with Repetitions Theorem. The number of different arrangements with repetitions (ordered selections where elements may repeat) of r elements taken from n is Remark. It is not necessarily that. Since we consider arrangements with repetitions, it should be that. 13

Example Suppose there are 6 traffic lights along some street. How many different arrangements of their signals exist, if each of them has exactly 3 signals {R,G,Y}? Solution: according to the arrangements with repetitions theorem there are arrangements. 14

Combinations with Repetitions Theorem. If repetition is allowed, the number of combinations (selections) of s elements that can be made from a set containing t distinct elements is Remark. It is not necessarily that. Since we consider combinations with repetitions, it should be that. 15

Example Suppose we want to buy 6 loafs of bread. How many different selections of bread exist, if there are two types of it {White, Brown}? Solution: according to the combinations with repetitions theorem there are combinations. 16

Homework Read Sections 8.2 and 8.4 paying a closer attention to examples. 17