ELEMENTARY COUNTING PROBLEMS LeAnhVinh University of Education Vietnam National University - Hanoi September 26, 2013 In this talk, we will discuss some fundamental counting techniques in which we follow closely [1, Chapter 3] Permutations Let start with a simple question We assume that n people arrived at a bank counter at the same time The bank teller will serve them one by one, so they must first decide the order in which they will be served How many different orders are possible? Certainly, there are n choices for the person who will be in the first position How many choices are there for the person who goes second? There are only n − choises as the person who went first will not go second, but everybody else can This observation implies that for each of n choices for the person to be served first, we have n − choices for the person to be served second We can then proceed in a similar manner: we have n − choices for the person to be served third, and so on, two choices for the person to be served next-to-last, and only one choice for the last person Therefore, we n · (n − 1) · · · · possible orders Definition 1.1 The arrangement of different objects into a linear order using each object exactly once is called a permutation of these objects The number n · (n − 1) · · · · of all permutations of n objects is called n factorial, and is denoted by n! 57 Note that by convention, 0! = One reason to see why we choose 0! to be rather than is the following problem Assume that we have n people ina room and m people in another room How many ways are there for people in the first room to form a line and people in the second room to form a line? The answer is clearly n! · m! In the special case, n = 0, then people in the second room can still form m! different lines Therefore, we will need n! = 0! = in this case The number n! is very important enumeration We have √ in combinatorial n the following approximation n! ∼ 2πn(n/e) The symbol n! ∼ α(n) means n! that α(n) → The above relation is called Stirling’s formula Example 1.2 Huong has five red balls, three yellow balls and two white balls to put into a row In how many different ways can she that? This problem differs from the above example in only one aspect: the objects are not all different This type of collection is called a multiset A linear order that contains all the elements of a multiset exactly once is called a multiset permutation Now, we can assign different numbers to the same color balls to have 10 distinct balls In this situation, we have 10! different ways to put 10 balls into a row Five red balls could be given five different labels in 5! ways Three yellow balls could be given three different lables in 3! ways and two white balls could be given three different lables in 2! Therefore, the labelling of all 10 balls can be done in 5! · 3! · 2! different ways once the balls are distributed Therefore, the number of ways that Huong can put the balls into a row is 10! 5! · 3! · 2! Similarly, we have the following general result Theorem 1.3 Let n, k, a1 , , ak be non-negative integers satisfying a1 + a2 + + ak = n Consider a multiset of n objects, in which objects are of type i, for all i ∈ [k] Then the number of ways to linearly order these objects is n! a1 ! · a2 ! · · ak ! 58 Strings over a finite alphabet Now are are going to study the question of how many times we can constructing strings, or words, from a finite sets of symbols, which we call a finite alphabet We will note require each occur a specific number of times, however we may require that each symbol occur at most once We first have the following easy result Theorem 2.1 The number of k-digit strings one can form over an n-element alphabet is nk Note that each digit can be chosen independently in n different ways so the totall number of choices is nk Using the above result, we have the following examples Example 2.2 The number of k-digit positive integers is · 10k−1 Solution The number of k-digit strings that can be made up from the alphabet {0, 1, , 9} is 10k However, not all these give a k-digit positive integer: those with first digit are not k-digit numbers Discard the first digit, we have a (k − 1)-digit strings Hence, the total number of k-digit positive integers is 10k − 10k−1 = · 10k−1 Example 2.3 The number of all subsets of an n-element set is 2n Solution For any A ⊂ [n], let f (A) be the string whose ith digit is if and only if i ∈ A and otherwise It is clear that A = B then f (A) = f (B) and for each string we have a corresponding subset Since the alphabet has only symbols, we have 2n subsets of n-elements Example 2.4 How many functions are there from [n] to [n] that are not one-to-one? Solution The totall of number of functions are from [n] to [n] is nn If f is one-to-one then {f (1), , f (n)} is a permutation of [n] This implies that there are total nn − n! functions satisfying the given condition Example 2.5 A city has ten recently built intersections Some of these will get traffice lights, and some of those that get traffice flights will also get a gas station In how many different ways can this happen? Solution Note that in each intersection, we have three possible choices: nothing is installed; traffic light; traffic light ad gas station Since these intersections are independent, the total off ways can be done is 310 59 Choice problems Now, we talk about the most interesting kind of elementary enumeration problems, the choice problems We have the following definition Definition 3.1 The number of k-element subset of [n] is denoted nk and is read n choose k The numbers nk are often called binomial coefficients To select a k-element subset of [n], we first select a k-element string in which the digits are distinct elements from [n] It is clear that we can it in n!/(n − k)! ways In chosen strings, the order of the elements does matter: each k-element subset occurs k! times among these strings as its elements can be permuted in k! different ways Therefore, the number of k element n! We note that 00 = 1, and it is convenient to define nk subsets is (n−k)!k! even in the case when n < k We define nk = in this case as no set has a subset that is larger than itself Example 3.2 A student work in a lab for six days in March She is not allowed to work two consecutive days in the lab In how many different ways can she choose the six days she will work in the lab? Solution Let x1 < x2 < x3 < x4 < x5 < x6 be the dates of the six days of March that the students will work at the lab Note that the requirement that there are no two consecutive numbers among the xi , and ≤ xi ≤ 31 for all i is equivalent to the ≤ x1 < x2 − < x3 − < x4 − < x5 − < x6 − ≤ 26 In other words, there is a bijection from the set of 6-element subsets of [31] containng no two consecutive elements and the set of 6-element subsets of [26], and we all know that there are 26 ways to that We also can apply the above trick when the chosen elements are allowed to be identical We have the following example Example 3.3 We play a lottery game where five number are drawn out of [99] Note that the numbers drawn are put back into the basket right after being selected To win the jackpot, one must have played the same multiset of numbers as the one drawn (regardless of the order in which the numbers were drawn) Solution Suppose that the drawn multiset is (x1 , x2 , x3 , x4 , x5 ) ≤ x1 ≤ x2 ≤ x3 ≤ x4 ≤ x5 ≤ 99 60 This requirement is equivalent to the following ≤ x1 < x2 + < x3 + < x4 + < x5 + ≤ 104 In other words, there is a bijection from the set of 5-elment multi-subsets of [99] and the set of 5-element subsets of [104 We know that there are 104 ways to that In general case, we have the following theorem Theorem 3.4 The number of k-element multisets whose elements all begong to [n] is n+k−1 k Exercises P 4.1 A company has 20 employees, 12 males and females How many ways we can form a group of employees that contains at least one male and at least one female? P 4.2 How many ways are there to list the digits {1, 2, 2, 3, 4, 5, 6} such that identical digits are not in consecutive positions P 4.3 How many ways are there to list the digits {1, 1, 2, 2, 3, 4, 5} such that the two 1s are in the consecutive positions? P 4.4 A student needs to spend six days in a lab during her last semester of studies After each day in the lab, she needs to spend at least six days in her office to analyze the data before she can return to the lab After the last day in the lab, she needs ten days to complete her report that is due at the end of the last day of the semester In how many ways can she this if we assume that the semester is of 15 weeks (105 days) P 4.5 How many ways are there to select an 11-member soccer team and a 5-member basketball team from a class of 45 students if a nobody can be on two teams b any number of students can be on both teams c at most one student can be on both teams? 61 P 4.6 We want to select as many subsets of [n] as possible so that any two selected subsets have at least one element in common What is the largest number of subsets we can select? P 4.7 ∗ A round robin chess tournament had 42 participants from two countries, 21 from each coutry There were no two players with the same number of points at the end Prove that there was at least one player who scored at least as many points against his compatriots as against the players of other country (Note that, a player gets one point for a win and one half of a point for draw.) P 4.8 ∗ A store has n different products for sale Each of them has a different price that is at least one dollar, or at most n dollars, and is a whole dollar A customer only has time to inspect k different products After doing so, she buys the product that has the lowest price among the k products she inspected dollars Prove that on average, she will pay n+1 k+1 P 4.9 ∗∗ A magic square is a squre matrix with non-negative integer entries in which all row sums and column sums are equal Let H3 (r) be the number of magic squares of size × in which each row and column have sum r Prove that r+4 r+3 r+2 H3 (r) = + + 4 P 4.10 ∗ a) At a round robin chess tournament, at least 3/4 of the games ended by a draw Prove that there were two players who had the same final score b) Now assume the tournament has been interrupted after t rounds, that is, after each player has finished t games (for simplicity, we assume that the number of player is even) Is it still true that if at least 3/4 of the games played ended by a draw, then there were two players with the same total score? c) Prove that if the games of the tournament are played in a random order (thre are no rounds - one player can play many games before another player starts), and the tournament is interrupted at some point Could it happen that 3/4 of the finished games ended by a draw but there were no two players with the same total score? d) Is there a constant K < such that if we organize the tournament as in the preceding case, and we interrupt the tournament at a point when at least K of the finished games ended by a draw, then there will always be two players with the same total score? 62 Hints and Solutions In this section, we will provide hints/solutions of odd problems in Section P 4.1 There are 20 ways to choose five people out of twenty employees 12 However, of these choices will result in male-only groups, and 85 will result in female-only groups Therefore, the number of good choices is 20 12 − − 5 P 4.3 Let us glue the two 1s together Then we only need to cpunt permutations of the multiset {1, 2, 2, 3, 4, 5} We know that we can it in 6! different ways 2! P 4.5 a) We have 45 choices for the soccer team Then we have to 11 choose from the remaining 34 people in 34 ways for the basketball team 45 34 Consequently, the final answer is 11 · b) Because we have no restriction at all, after choosing the soccer team, we can choose the basket ball team in 45 ways So the total number of 45 45 choices is 11 · c) All compositions in the first part also satisfy the condition Now, we only need to count the number of compositions in which there is exactly one student on both teams We have 45 choices for this person, then 44 · 344 10 ways to choose the remaining players from the rest of the class Thus, the total number of possibilities is 45 34 44 · + 45 · · 344 11 10 P 4.7 Let A be the country whose players scored at most as many points in the international games as players from the other country Take the 21 players from A, and let a1 , a2 , , a21 denote the number of points they get againts their countrymen Let b1 , b2 , , b21 denote the number of points they get againts players from the country B Now we assume that the claim is false, i.e < bi for all i It means ≤ bi − 0.5 for all i Summing these inequality over all i ∈ [21], we have 21 21 ≤ i=1 bi − 21/2 i=1 63 On the other hand, 21 i=1 = 21·20/2 as any two players from A played each other once, and in each of those games, one point was given to the players This implies that 21 212 21 · 20 21 + = bi ≥ 2 i=1 Since players from A got at most half of all points from international games, we have 21 212 bi ≤ i=1 21 21 Puting the above inequalities together, we see that i=1 bi = This implies that = bi − 0.5 for all i, or the total score of the ith player from country A was + bi = 2ai + 0.5 which is never an integer Therefore, no player from country A has a final score that is an integer By the same argument, no player from country B has a final score that is an integer This is a contradiction as we know there are no two players with the same final score The number of possible non-integer final scores is less than 42 are 0.5, 1.5, , 40.5, which is only 41 different scores for 42 players So, there must be a player who did better against his compatriots than against players from the other country P 4.9.It is clear that three elements on the diagonal and the second on the first row determine all the rest of the magic square Let a, b, c be the numbers on the main diagonal and d be the second number on the first row The obtaining square is a magic square if and only if the following inequalities holds: a+d b+d c c a+d+b−c ≤ ≤ ≤ ≤ ≤ r r a+d b+d r We have three cases • Suppose ≤ a ≤ b and ≤ a ≤ c One can check that the following conditions are equivalent to a ≤ 2a + d − c ≤ a + b + d − c ≤ b + d ≤ r 64 Hence, we have r+4 magic squares in this case • Suppose that a > b and c ≥ b One can check that the following conditions are equivalent to b ≤ 2b + d − c ≤ a + b + d − c − ≤ a + d − ≤ r − Hence, we have r+3 magic squares in this case • Suppose that a > c and b > c One can check that the following conditions are equivalent to c ≤ b − ≤ b + d − ≤ a + b + d − c − ≤ r − Hence, we have r+2 magic squares in this case Thus, the number of × magic squares of line sum r is H3 (r) = r+4 r+3 r+2 + + 4 Furthermore, the three terms in this sum count the magic squares in which the (first) minimal element of the main diagonal is the first, second, or third element References [1] M B´ona, A walk through combinatorics: An introduction to enumeration and graph theory, World Scientific Publishing, 2nd Ed 2006 65 ... have 45 choices for this person, then 44 · 344 10 ways to choose the remaining players from the rest of the class Thus, the total number of possibilities is 45 34 44 · + 45 · · 344 11 10 P 4. 7... than 42 are 0.5, 1.5, , 40 .5, which is only 41 different scores for 42 players So, there must be a player who did better against his compatriots than against players from the other country P 4. 9.It... − − 5 P 4. 3 Let us glue the two 1s together Then we only need to cpunt permutations of the multiset {1, 2, 2, 3, 4, 5} We know that we can it in 6! different ways 2! P 4. 5 a) We have 45 choices