LESSON 1: INTRODUCTION TO COUNTING

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1 LESSON 1: INTRODUCTION TO COUNTING Counting problems usually refer to problems whose question begins with How many. Some of these problems may be very simple, others quite difficult. Throughout this course we will see some of the numerous methods used in solving such problems. We will start with the easiest ones. 1. Finding a counter Often we need to know how many terms a given sequence contains. Then identifying a general expression of the terms is very useful. Examples: 1. How many positive integers are there from n to m (n < m)? Solution: Writing the numbers n, n + 1, n + 2,..., m of the sequence as (n 1) + 1, (n 1) + 2,..., (n 1) + (m n + 1), we see that all the numbers are of the form (n 1) + k where the counter k goes from 1 to m n + 1. As k takes m n + 1 values, there are m n + 1 numbers in the sequence. 2. How many terms do the sequences below contain: (a) 0, 5, 10, 15,..., 2010 (b) 201, 204, 207,..., 2010 (c) 0, 2, 6, 12, 20,..., 9900? Solution: It is useful to emphasize a counter. (a) The terms of the sequence can be written as 5 0, 5 1, 5 2,..., Then it is easy to see that the terms of the sequence are of the form 5n where n counts from 0 to 402. Therefore the sequence has 403 terms. (b) The numbers can be written as 3 67, 3 68,..., 3 670, that is 3k with the counter k ranging from 67 to 670. According to Example 1, the sequence contains = 604 terms. (c) The terms of the sequence are 0 1, 1 2, 2 3,..., We can take as counter either the first factor of each term (it will range from 0 to 99), or the second factor (ranges from 1 to 100). In both cases it is clear that the sequence has 100 terms. 3. The increasing sequence 1, 3, 4, 9, 10, 12, consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the 100 th term of this sequence. (AIME 1986 and also TST Puerto Rico 2012) 1

2 Solution: Since the base 3 representation of a number is obtained by writing the number as a sum of powers of 3, the statement of the problem suggests considering the base 3 representation of the numbers of the sequence. Written in base 3, the sequence becomes: 1, 10, 11, 100, 101, 110,.... This is none other then the sequence of the positive integers written in base 2. The 100-th term of this second sequence is the base 2 representation of the number 100. Since 100 = = , we obtain this number to be Then the 100-th term of the initial sequence is = Rule of the sum The rule of sum or addition principle is a basic counting principle. The idea is that if we have a ways of doing something and b ways of doing another thing and we can not do both at the same time, then there are a + b ways to choose one of the actions. More formally, the rule of sum is a fact about set theory. For two disjoint sets, A and B, it states that A B = A + B. More generally, for a finite collection of pairwise disjoint sets it states that sum of the sizes of these sets is the size of the union of the sets: if S 1, S 2,..., S n are pairwise disjoint sets, then we have: S 1 + S S n = S 1 S 2 S n. This rule is often applied when treating cases. ) Examples: 1. How many positive integers are smaller than 2011 and have an even number of digits? Solution: There are two cases: the case of the 2 digit numbers (all of them are less than 2011) and the case of the 4 digit numbers that are smaller than If A = {10, 11,..., 99} and B = {1000, 1001,..., 2010}, then A = 90, B = 1011 and the set of the numbers satisfying the conditions of the problem, A B, has A B = A + B = = 1101 elements. 2. How many three digit integers are either even or multiples of 5? Solution: A number that is even or a multiple of 5 has as its last digit one of the digits 0,2,4,5,6, or 8. If we denote by S 1 the set of the three digit integers that have the last digit 0, by S 2 the set of those that have the last digit 2, and so on, S 6 the set 2

3 of the 3 digit integers that have the last digit 8, then the question of the problem is to find S 1 S 2... S 6. As these sets are mutually disjoint, applying the rule of the sum gives S 1 S 2... S 6 = S 1 + S S 6. S 1 = {100, 110,...990} has 90 elements (the counter goes from 10 to 99), therefore S 1 = 90. Similarly, S 2 = S 3 =... = S 6 = 90 and S 1 S 2... S 6 = 6 90 = How many squares, of all possible sizes, can one count on an 8 8 chessboard? Solution: On the chessboard there are squares 1 1, 2 2,..., 8 8. According to the rule of the sum, we need to count the squares of each size and then add up the results obtained in each case. Obviously there are squares. In order to count the 2 2 squares we identify each square by a distinctive point, namely the upper left corner of each square. (The center of each square would have been another natural option.) The borderlines of the 64 unit squares determine a grid of 9 horizontal and 9 vertical lines. These lines intersect in 81 points. Which of these 81 points can be the upper left corner of a 2 2 square? Answer: those situated on the first 7 horizontal lines and on the first 7 vertical lines. Therefore there are 7 7 such points, that is there are squares. We can count the 3 3 squares similarly obtaining 6 2 such squares. Continuing in the same manner, we find 5 2 squares of size 4, 4 2 squares of size 5, 3 2 squares of size 6, 2 2 squares of size 7 and the whole 8 8 board. In total, there are (8 + 1)( ) = = squares on the chessboard. We will see later on how to count the size of the union of some, not necessarily disjoint, sets of known size. For now, let us simply note, that A B = (A\B) B = A \ B + B (one can MAKE the two sets to be disjoint). Example: How many two digit numbers contain the digit 1 in their decimal representation? Solution: There are two natural cases to consider: the numbers that have 1 as their first digit, and the numbers that have 1 as their second digit. But in this way the two cases are not disjoint: it is possible that a number has both its digits equal to 1 (namely 11). Therefore the rule of the sum does not apply for these two cases. Applying the rule of the sum without checking that the sets are disjoints is one of the main reasons of overcounting. Here, in order to be able to apply the rule of the sum, we need to make the two cases disjoint. To this effect, consider the case of the numbers that have 1 as their first digit and, for the second case, the numbers that have 1 as their second digit BUT NOT as their first digit too. Then in the first case we obtain the numbers from A = {10, 11,..., 19}, and in the second 3

4 case the numbers belonging to B = {21, 31,..., 91}. Then, by the rule of the sum, A B = A + B = = Rule of the product The rule of product is another intuitive principle stating that if there are a ways to do something and b ways to do another thing, then there are a b ways to do both things. This rule only applies when the choices of the two actions are independent. The rule can be extended to any finite succession of independent actions (choices): the total number of possible outcomes is equal to the product of the number of outcomes for each individual action. Examples: 1. When you decide to order pizza, you must first choose the type of crust: thin or deep dish (2 choices). Next, you choose the topping: cheese, pepperoni, or sausage (3 choices). Using the rule of product, you know that there are 2 3 = 6 possible combinations of ordering a pizza. 2. When counting how many two digit numbers are multiples of 3, it is obvious that the first digit can be chosen in 9 ways and the second digit in 10 ways. But these choices are not independent. If you begin by choosing the first digit, the choice of the second one depends on the choice of the first digit. The rule of the product does not apply. 3. In how many different ways we can assign offices to two employees if the total number of offices in the company is 100? Solution: The first employee has the choice among 100 offices. Once he has chosen his office, the second employee has to choose from the remaining 99 offices. By the rule of the product, there are = 9900 ways of assigning the offices. 4. In how many different ways can one arrange 5 books on a shelf? Solution: There are 5 positions on the shelf, each of them eagerly awaiting a book. The first book chooses one of the 5 positions. Then, the second book has to choose from the remaining 4 positions. The third book has 3 options left, then the forth book chooses from the two available positions, and finally, the last book will occupy the last empty place. By the rule of the product, there are = 120 ways of arranging the books. 4

5 5. In how many ways can one distribute m (distinguishable) books to n persons? Solution: If each book is to choose its master, the first book has n possible choices, the second book also n choices (it is allowed to choose the same person as the first book), and so on, each of the m books having n choices. By the rule of the product, there are n } n {{... n} = n m ways to distribute the books. m times 6. If the prime factorization of a positive integer N is N = p a 1 1 p a p a k k, then the number of positive divisors of N is given by d(n) = (a 1 +1) (a 2 +1)... (a k +1). Solution: The prime factorization of any positive divisor D of N is D = p b 1 1 p b 2 where the (distinct) primes p 1, p 2,..., p n are those from the prime factorization p b k k 2... of N, while the powers b 1, b 2,..., b n satisfy 0 b j a j, j = 1, 2,..., n. There are as many divisors as there are successions of choices for the powers b j. But as b j {0, 1, 2,..., a j }, there are a j + 1 choices for each b j. By the rule of the product, there are (a 1 + 1) (a 2 + 1)... (a k + 1) combined choices, and the same number of positive divisors. 7. An ordered pair (m, n) of non-negative integers is called simple if the addition m + n in base 10 requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to (AIME 1987) Solution: Let us write m = (abcd) and n = (efgh) even if m or n do not have 4 digits but less (e.g. we will write 0032 instead of 32). If (m, n) is simple, then a+e = 1, b+f = 4, c+g = 9 and d+h = 2. a+e = 1 means (a, e) {(0, 1), (1, 0)} (2 options), b + f = 4 means (b, f) {(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)} (5 possibilities). Similarly, c + g = 9 has 10 possibilities and d + h = 2 has 3 possibilities. Since all these options are independent, we have = 300 simple pairs. 8. How many different bit strings of length 20 are there? How many of these strings are symmetric? Solution: Each bit can take two possible values: 0 or 1. Therefore there are 2 } 2 {{... 2} = 2 20 such strings. 20 times The fact that a string is symmetric means that if we choose a certain value for the first bit, the value of the last bit is imposed (must be the same as the first one). Similarly, the second last has to be the same as the second one, so we only have the liberty to choose the second bit, the other one is fixed by the choice of its symmetric. Generally, we can only choose the first 10 bits, the choice of the other ones being uniquely determined. Thus there are 2 10 symmetric strings. 5

6 9. (a). Given an n-element set S, in how many ways can we choose a subset X of S? (b). Given an n-element set S, in how many ways can we choose an ordered pair (X, Y ) of subsets of S? (c). Given an n-element set S, in how many ways can we choose an ordered pair (X, Y ) of subsets of S such that X Y = S? (d). Given an n-element set S, in how many ways can we choose an unordered pair {X, Y } 1 of subsets of S such that X Y = S? (e). Given an n-element set S, in how many ways can we choose an ordered triple (X, Y, Z) of subsets of S such that X Y Z = S? (f). Given an n-element set S, in how many ways we can choose an ordered triple (X, Y, Z) of pairwise disjoint sets such that X Y Z = S? Solution: (a) Each of the n elements of S has two options: it can choose either to belong to X, or it can choose not to belong to X. Thus there are 2 } 2 {{... 2} = 2 n ways of choosing a subset of S. n times Important to remember: an n-element set has 2 n subsets. (b) We can count in two ways: each element can belong to: only X, only Y, both X and Y, neither X nor Y, so it has 4 possibilities. Then there are 4 n such ordered pairs; or each of the two sets can be chosen in 2 n ways, so there are 2 n 2 n = 4 n such pairs. (c) Each element of S can: belong only to X, only to Y, both X and Y, so it has 3 possibilities. Then there are 3 n such pairs. (d) To every pair of ordered pairs (X, Y ) and (Y, X) with X Y there is an associated unordered pair {X, Y }. Between the ordered pairs from (c) only the pair (S, S) is not of the form (X, Y ) with X Y. So 3 n 1 of the ordered pairs from (c) come in pairs (X, Y ), (Y, X). To these ordered pairs correspond 3n 1 unordered 2 pairs. And we need to add the singleton {S} that comes from the ordered pair (S, S). We get 3n = 3n + 1 unordered pairs. 2 2 (e) Each element of S has 7 possible destinations: it can belong to: X, Y and Z, or only to X and Y, or only to Y and Z, or to only Z and X, or only to X, or only to Y, or only to Z. Then there are 7 n such ordered triples. (f) Each element has three choices: it can belong to exactly one of the sets X, Y and Z, so there are 3 n such sets. 1 where X and Y need not be distinct 6

7 10. (a) How many functions are there from a set with m elements to a set with n elements? (b) How many injective (one-to-one) functions are there from a set with m elements to a set with n elements? Solution: (a) Each of the m elements has n possible destinations, therefore there are n m functions. (see Example 5.) (b) Since any two elements of the first set have to go to different elements of the second set, the second set needs to have at least m elements, otherwise there are no such injective functions. If n m, the first element has n possible destinations, then the second element has n 1 destinations, the third n 2 destinations and finally, the last element, the m-th, has n m + 1 possible destinations. We obtain that there are n(n 1)(n 2)... (n m + 1) such functions. Remark: If we denote N! = N and define 0! = 1, then the result above n! can be written as (n m)!. For n = m = 5 we obtain the situation from Example Each unit square of a 1 n rectangle is painted with one of four different colors such that no two neighboring squares have the same color. (a) Find the number of distinct symmetrical rectangles that we can obtain. (b) Find the number of distinct rectangles that satisfy the condition that any three consecutive squares are painted with three different colors. Solution: (a) The answer depends on whether n is even or odd. If n is odd, n = 2k + 1, then we can choose the color of the first square in 4 ways and the colors of all the other squares until square number k + 1 in 3 ways. The colors of the other squares are determined uniquely by symmetry. There are }{{} k times = 4 3 k such rectangles. If n is even, we look at the two central squares. Because of the symmetry they need to have the same color, so in this case we must have two neighboring squares with the same color. Therefore, if n is even, there are no such rectangles. (b) The color of the first square can be chosen in 4 ways. The color of the second square in 3 ways (different from the color of the first cell). The color of all the other cells can be chosen in 2 ways (to be different of the colors of the two preceding squares). There are n 2 = 3 2 n such rectangles. 12. An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there? (AIME 2003) 7

8 Solution: Let us consider cases according to the common sum of the two leftmost and rightmost digits. If it is 1, the first two digits must be 10, the last two can be either 01 or 10. Therefore there are 1 2 possibilities for the common sum to be 1. If the common sum is 2, the first two digits can be 11 or 20, the last two digits 02, 11 or 20. Thus there are 2 3 possibilities with the common sum equal to 2. Similarly we obtain 3 4 possibilities with the sum 3, and so on, until we obtain 9 10 possibilities with the sum 9. Then, for the sum 10 we obtain, for both the first two digits and the last two digits, 9 options: 19, 28, 37, 46, 55, 64, 73, 82, 91, therefore 9 9 possibilities. Then, for the sum 11 there will be 8 8 possibilities and so on, until we get, for the sum 18, only 1 1 possibilities (namely 9999). In total we have = = 9(9 + 1) (9 + 1)( ) 6 = = The numbers 1447, 1005, and 1231 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. How many such numbers are there? (AIME 1983) Solution: There are two types of such numbers: those that repeat the digit 1, and those that repeat another digit. Those with two 1-s can be (11xy), (1x1y) or (1xy1) where x, y are distinct digits, different from 1. For each of the 3 situations, x can be chosen in 9 ways and then y in 8 ways, so there are = 216 numbers with two 1-s. With the repeated digit other than 1, we have numbers of the following shapes: (1aab), (1aba) and (1baa), where in all the cases the digits a and b must be distinct and different from 1. For each of the 3 situations, a can be chosen in 9 ways and then b in 8 ways, so there are = 216 numbers that repeat some other digit than 1. In total there are = 432 convenient numbers. 14. For {1, 2, 3,..., n} and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for {1, 2, 4, 6, 9} is = 6 and for {5} it is simply 5. Find the sum of all such alternating sums. (AIME 1983) Solution: Before solving the problem for a general n, consider some easier cases. Start with n = 2. The 3 alternating sums are 1, 2 and 2 1 and their sum is 2 2 = 4 because the 1-s cancel. For n = 3, the alternating sums are 1, 2, 3, 2 1, 3 1, 3 2, and their sum is 4 3 = 12 because the 1-s and the 2-s cancel. This suggests that even in the general case, somehow all the 1-s, 2-s,..., (n 1)-s will cancel and since there are 2 n 1 sets that contain the number n, the final result 8

9 will be 2 n 1 n. (Note that in each subset that contains n, n will be the largest element of the subset, so in the alternating sum it will have the coefficient +1.) Consider the 2 n 1 subsets of {1, 2,..., n 1}. To each such subset A we can add the element n obtaining the set A {n}. All the subsets of {1, 2,..., n} are either subsets of {1, 2,..., n 1}, or sets of the form A {n} with A a subset of {1, 2,..., n 1}. This allows us to group (all) the subsets of {1, 2,..., n} into pairs ( A, A {n} ), where A is a subset of {1, 2,..., n 1}. Now let us compare the alternating sums for A and A {n}. Except for n, they will contain the same terms, only with opposite signs. For example, if A = {1, 3, 7}, the alternating sum is , while for A {n} it is n When we add the alternating sums of A and A {n}, all the terms will cancel, except for n, so the sum of the two alternating sums from each pair is n. Since we have 2 n 1 pairs, the final result is n 2 n 1. (Since we are supposed to only consider non-empty sets, from the pair (, {n}) we should only consider the alternating sum of {n} but this is also n.) PRACTICE PROBLEMS 1. Bart has $3 and $7 dollar bills only. In how many different ways can he give $40 to Lisa using just his money? Answer: Two ways: 40 = = Determine the number of ways in which we can arrange four red and six green marbles in a row so that all green marbles are next to each other. Solution: Since the green marbles can not be separated consider them as one. This group can be placed in front of the first red marble, between the first and the second red marble, between the second and third, between the third and the forth or, finally, behind the forth red marble. So there are 5 possibilities. 3. A wooden cube with edge length n units (where n is an integer > 2) is painted in black all over. By slices parallel to its faces, the cube is cut into n 3 cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is n? (AHSME 1985) (A) 5 (B) 6 (C) 7 (D) 8 (E) none of these Solution: Let us count first the cubs with exactly one face painted black. On each of the 6 faces of the wooden cube, there are n n faces of little cubes that are painted black. The frame belongs to cubs that have other faces painted black 9

10 as well. This leaves us with (n 2) (n 2) faces that belong to cubs that do not have any other painted face. Therefore we have 6(n 2) 2 such cubs. If we remove the painted cubs (the first and last layer from front, rear, top, bottom, left and right), we obtain an (n 2) (n 2) (n 2) cube that is free of paint. It contains (n 2) 3 little cubs. So n must satisfy the condition 6(n 2) 2 = (n 2) 3, therefore n = How many bit strings of length 8 either start with 1 or end with the two bits 00? Solution: There are two types of convenient strings: Those that start with 1, and those that end with 00. However, we can not apply directly the rule of the sum because the two types are not disjoint: the strings starting with 1 and ending with 00 would be counted twice. To make the two cases disjoint, consider the strings that start with 1, then the strings that start with 0 and end with 00. There are 2 7 strings of the first type (each of the last 7 bits can be chosen in 2 ways), and 2 5 strings of the second type (each of the remaining 5 bits can be chosen in 2 ways). Thus there are = 160 such strings. 5. Determine the number of positive integers whose decimal representation contains exactly 4 digits, each of which is distinct and none of which is the digit 3. Solution: The first digit may not be 0 or 3, so it can be chosen in 8 ways. Then the second digit may not be 3, and can not be equal to the first digit, so there are 8 ways to choose it. Similarly, there are 7 ways of choosing the third digit and 6 ways for the last digit. By the rule of the product, there are = 2688 numbers with the desired condition. 6. Define a good word as a sequence of letters that consists only of the letters A, B, and C - some of these letters may not appear in the sequence - and in which A is never immediately followed by B, B is never immediately followed by C, and C is never immediately followed by A. How many seven-letter good words are there? Solution: The first letter can be chosen in 3 ways, all the others in 2 ways, therefore the result is = Let (s 1, s 2,..., s n ) be a permutation of the numbers 1, 2,..., n. For how many of these permutations is true that s k k 2 for each k = 1, 2,..., n? (Náboj Competition, 2009) Solution: First, we must choose s n {n 2, n 1, n} (3 options). Then, s n 1 {n 3, n 2, n 1, n} (after having chosen s n {n 2, n 1, n}, there are 3 remaining options for s n 1 ). Then we chose s n 2 {n 4, n 3, n 2, n 1, n} 10

11 (3 remaining options) and so on. We will always have 3 options for choosing s n, s n 1,..., s 4. For s 3, s 2 and s 1 the condition s k k 1 is always fulfilled so we can asign them in any possible way to the 3 remaining values. This can be done in 3! ways, so the answer is 3 n 3 3! = 2 3 n A (2n + 1) (2n + 1) square is cut into (2n + 1) 2 unit squares. How many squares, of all possible sizes, have been formed? How many of them contain the unit square in the center? 11

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