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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 the corner cells is told that he will be released, provided he can get into the diagonally opposite cell after passing through every other cell exactly once Can the prisoner obtain his freedom? Solution Count the number of the walls the prisoner has to cross to reach to the corner cell diagonally opposite to the beginning corner cell after passing through every other cells exactly once Then count number for walls separating those two cells (EVEN or ODD? 1 Read the algorithm of de la Loubère s method in the book 2 Section 24 Proof Suppose that 100 numbers are chosen We write each number a = 2 k a 0 with a 0 odd, which is called the odd part of a If two numbers from the selection have the same odd part, then one of the two will divide the other one and we are done Otherwise, all 100 numbers have distinct odd parts, in particular all odd numbers from 1 up to 199 are odd parts of the 100 selected numbers Let a be one of 100 selected number such that a 15 Write a = 2 k a 0 with a 0 odd Thus 0 k If k = 0, then a = a 0 is odd and, a, 5a, 7a, 9a, 11a, 1a are odd numbers no larger than 199 So they are odd parts of some selected numbers Hence, a divides at least selected numbers If 2 k > 0, then a 0 < Thus, a 0 7 In this case, a 0, 9a 0 and 27a 0 are odd and less than 200 (27 7 = 19, thus they are the odd parts of some of the 100 selected numbers, say that they are b 1 = 2 e 1 a 0, b 2 = 2 e 2 9a 0, and b = 2 e 27a 0 If one of e 1, e 2, e is at least 2, then a divides one of numbers b 1, b 2 b Otherwise two of e 1, e 2, e equal by the Pigeonhole principle Then two of b 1, b 2, b with equal powers of 2 will have one divide the other If k =, then a 0 < 2, thus a 0 = 1 and a 0, 9a 0, 27a 0, 1a 0 are odd and less than 200, thus are odd parts of some of the 100 selected numbers, say that they are b 1 = 2 e 1 a 0, b 2 = 2 e 2 9a 0, b = 2 e 27a 0, and b 4 = 2 e 4 1a 0 If one of the four powers of 2 is at least, then a divides the corresponding number Otherwise, 0 e i 2 for i = 1, 2,, 4 Pigeonhole principle implies that two of the powers of 2 in the four numbers b 1, b 2, b, b 4 have to be the same That gives two numbers such that one divides the other A in application 5, if the pigeonhole fails, then 100 numbers have 100 distinct odd parts Thus the numbers 2 e a 0, 2 e 5 5a 0, 2 e 7 7a 0, 2 e 9 9a 0, 2 e 11 11a 0, 2 e 1 1a 0 are in the selection in addition to a for some e, e 5, e 7, e 9, e 11, e 1 If any one of them is at least 2, then a divides the corresponding number, otherwise they are all at most 1 and two of them have to equal Hence those corresponding number will have one divides the other Choose n + 1 numbers then follow the argument almost identically as in Application 5 5 Line up the n + 1 number from small to the largest then add up the consecutive differences 7 Each integer n can be written as n = 100q + r with r 50 1

9 For each group of people, the group age is the sum of the ages of people in that group There are totally 2 10 = 1024 different groups, while the group ages have to maximal of 10 0 = 00 Two groups must have the same group age But the two group could share a common member Through away the people from both groups if they are in both groups 10 Follow the application 4 15 Prove that, for any n + 1 integers a 1, a 2,, a n, a n+1, there exist two of the integers a i and a j with i j such that a i a j is divisible by n Proof Hint: Consider the n boxes labeled by 0, 1,, n 1 Place the integers a i in the box labeled r is the remainder of a i is r when divided by n Pigeonhole principle implies that there is a box with two difference objects a i and g j with i j, ie, a i and a j have the same remainder when divided by n Hence the difference of these two numbers have the remainder 0 when divided, ie, a i a j is divisible by n 17 Write a i = 2m i to be the number of acquaintances of the ith person Then 0 m i 49 If each number appears exactly twice, then there will be two people who don t know any one while there is one person who knows 9 people not including him/herself That is impossible 19 (a Divide the triangle into four (1 + equilateral triangles with sides length 1/2 (b Divide the triangle into nine (1 + + 5 equilateral triangles with sides length 1/ (c Divide the triangle into (1 + + 5 + + (2n 1 equilateral triangles with sides length 1/n 1 easy 5 Count the number of the factors 2 and 5 Section In how many ways can six men and six ladies be seated at a round table if the men and ladies are to sit in alternate seats? Solution The arrangement can be achieved through two stages 9 Count the number of the situations that A and B sit next to each other by considering AB and BA as single person For the send part, consider the case when AB occurs 10 A committee of 5 is to be chosen from a club that boast a membership of 10 men and 12 women How many ways can the committee be formed if it has to have at least 2 women? How many ways if, in addition, one particular man and one particular woman who are members of the club, refuse to serve together on the committee? Solution The total number of possible committees can be formed without any restrict is ( 22 5 Among them, the number of committees without any women is ( 10 5 and the number of the committee with exactly one woman is 12 (10 4 Thus the number of committees with at least two women is ( ( 22 5 10 ( 5 12 10 4 If A is the particular man and B is the particular woman who can not serve together Then the total number of committees with both A and B on the committee is ( 20 without any other restrictions But the number of such committees with exactly one woman is ( ( 9 Thus 20 ( 9 is the number of committees with at least two women and A and B are both on This number should be excluded from the earlier answer Thus the number of committees with at least two women and not A and B both on the committee is ( ( 22 5 10 ( 5 12 10 ( 4 ( 20 ( 9 2

11 How many sets of numbers each can be formed from the numbers {1, 2,,, 20} if no 2 consecutive numbers are to be in a set Solution First count the total number of number subsets regardless two consecutive numbers being in ( a set or not This number is the number of the -combinations of the set of 20 elements, 20 which is Then count the number of -element subsets with exactly one pair of consecutive numbers in it For each pair {i, i + 1} for 2 i 1, there 20 4 = 1 ways to add the third number that won t make new consecutive numbers with two already chosen, ie, any number other than {i 1, i, i + 1, i + 2} For i = 1 or 1 = 19, there are 20 = 17 choices So the total is 17 1 + 2 17 = 17 1 We still need to count the number of -number subsets that contain -consecutive numbers There are ( 1 of them ({i, i + 1, i + 2} for i = 1,, 1 Finally the total 20 number of -number subsets is 17 1 1 = 1140 24 = 1 12 A football team of 11 players is to be selected from a set of players, 5 of whom can play only in the backfield, of them can only play on the line, 2 of them can play either in the backfield, or on the line Assuming a football team has 7 men on the line and 4 men in the backfield, determine the number of football teams possible Solution We divide the formation of the team into the following situations: (1 None of those two players (A and B who can player in the backfield or on the line is on the team In this case, there are ( ( 7 5 4 = 40 (2 Exactly one of A and B is on the team There two choices For each choices, there are two situations: (a This person plays in the back field and there are ( 5 ( 7 = 10 many possible (b This person is plays on the line, there are ( 5 4( = 5 2 = 140 possible teams Thus the total number of teams when only one of A and B plays is 2(10 + 140 = 00 ( Both A and B in the team Now there are four different cases (a Both A and B are on line, there are ( ( 5 4 5 ways (b Both A and B are in the back field, there are ( 5 ( 2 7 ways (c One of the A and B is on the line and the other is in the backfield There are 2 (5 ( ways Finally the total number of ways is the sum of the numbers above 14 A classroom has two rows of seats each There are 14 students, 5 of whom always sit in the front row and 4 of whom always sit in the back row In how many ways can the students be seated Solution We will regard the eight seats in each row different The arrangement process can be divided into three stages 1 Let the 5 students who always sit in the font row choose their seats There are P (, 5 possible outcomes 2 Let the 4 students who always sit in the back row choose their seats There are P (, 4 possible outcomes Let the remaining five students to fill choose their seats from the remaining 1 9 seats There are P (7, 5 different ways By the multiplication principle, the total number of ways to seat the students is P (, 5 P (, 4 P (7, 5 15 (a Select 15 women out of the 20 then distribute them to 15 men

(b First choose 10 men then choose 10 women to match 1 In how many ways can 2 red and 4 blue rooks be placed on an -by- board so that no two rooks can attach on another Solution Note that there cannot be more than one rook in any column or row in order to avoid any rook attacking any other rooks The placement can be done to two stages ( Stage 1 Place 2 red rooks Ch oose 2 rows out of rows and there are ways Then 2 for ( each place the two rooks in the two rows chosen and there are 7 of choices Thus there are 7 ways to place 2 red rooks 2 Stage 2 For each placement of the two ( red rooks, choose 4 rows out the six remaining rows to place the four blue rooks There are ways Then place the four rooks in the remaining 4 ( columns for each choice of the four rows There 5 4 ways to do so Thus there are 5 4 ways to place the four blue rooks in the remaining board 4 ( 72 ( 5 4 2 By multiplication principle, there are ways to place the six rooks on the 2! 4! board so that no two rooks attack one another Another way: First place indistinguishable rooks on an -board Then choose 2 of them to be painted red 19 (a Choose 5 rows for the red rooks others are for blue rooks Decide the column positions! ways of doing so (b Choose 5 rows for red rooks, then choose from the remain rows (7 of them choose three for blue rooks Then chose columns to place the rooks standing in different rows 2 Determine the number of 11-permutations of the multiset { a, 4 b, 5 c} Solution There are three types of objects in the multiset S which has size 12 To get 11- permutation, one of the objects has be left behind There are three different ways to leave one object behind, that is, either an a, or a b, or a c The number of 11-permutations with 2 a s is 11! 2! 4! 5! ; The number of 11-permutations with b s is 11!!! 5! ; The number of 11-permutations with 4 c s is 11!! 4! 4! ; Thus the total number of 11-permutations is 11! 2! 4! 5! + 11!!! 5! + 11!! 4! 4! There twelve objects in the multiset decide which one of the object to be unselected 7 A bakery sells different kinds of pastry If the bakery has at least a dozen of each kind, how many different options for a box of dozen pastry are there? What if each box contains at least one of each kind? Solution Let x 1, x 2, x, x 4, x 5, x be, respectively, the numbers of the first, second, third, fourth, fifth, and sixth pastry in a dozen Then each x i is a non-negative integer and x 1 + x 2 + x + x 4 + x 5 + x = 12 This becomes a question of 12-combinations of a multiset with six types of objects and infinite repetition numbers Then total number is 4 ( 12 + 1 12

If there is at least one of each kind in the ( dozen, then each x i 1 Set y i = x i 1 Then + 1 y 1 + y 2 + y + y 4 + y 5 + y = Thus there are 41 In how many ways can 12 indistinguishable apples and 1 orange be distributed among three children in such a way that each child gets at least one piece of fruit? Solution Since each child should get at least one piece of fruit and all fruits have to be distributed, do the distribution in two stages Stage 1 Choose one of the three children to get the orange and the other two for one apply each There are ways Stage 2 For each outcome of the stage 1, distribute the remaining 10 apples to the three children Each child has the ability to take as many as he/she is given This becomes problem of 10-combinations of a multiset with three types of objects (apples received by each ( child with 10 + 1 infinite repetition numbers (each child can take as many as given Thus there are 10 ways ( 10 + 1 Finally, multiply the ways from two stages to get the total number of ways 10 distributing the fruits to three children such that each child gets at least one piece of fruit 42 Distribute the lemon and the lime drinks first then give each of the remaining two children an orange drink Then distribute the rest of orange drinks without any restrictions (combinations of multiset of the number of the integral solutions of an equation with conditions Section 5 Expand the binomial with x and 2y as two variables first 7 Consider x = 1 and y = 2 in the binomial theorem 9 Set x = 1 and y = 10 binomial theorem 11 The left hand side is the number of the k-combinations that has at least one of a, b, c in it For( the right have side, n 1 k 1 is the number k-combinations of S with a in them; ( n 2 k 1 is the number of k-combinations without a but has b in them; ( n k 1 is the number of k-combinations without a and b but has c in them; 15 Consider the binomial expansion of the function f(x = (1 + x n and compute f (x on both sides Then evaluate f ( 1 with n 1 > 0 21 Take all negative signs out from ( r k = ( r( r 1 ( r k+1 and then reverse the order of the k! product of the numerator 2 (a The paths are determined by when to walk in an east-west direction street The number of paths is ( 24 10 (b ( ( 9 4 ways to walk to her friends home There are 15 5 ways from her friends home to school Total number ways from her home to school passing through her friend s home is ( 9 15 4( 5 (c There are ( 9 9 4( ( ways 25 Consider a group people with m 1 women and m 2 men To choose n-people from the group, consider the the number k of women chosen 5

1 Note that the clutter of size at least con not have 0-combination nor 4-combination Then show that there can not be any -combinations in it Otherwise, say {1, 2, } is in the clutter of size Among the total 2 4 = 1 combinations, only 14 are usable and seven of them are subsets of {1, 2, } Seven of the remaining 9 subsets, all contain 4 in them and {4} cannot in the clutter nor can any three combination with 4 in them (which will have four subsets with 4 in them leave not enough subset for subsets to form a clutter By duality, there cannot be any 1-combination in the clutter 40 Use multinomial coefficients 41 Use multinomial expansion 4 Consider the n-permutations of the multiset S = {n 1 a 1,, n t a t } Counting those starting with a 1, those starting with a 2, etc Section 7 1 Use inclusion-exclusion principle by considering the numbers that are divisible by 4, 5, or respectively and their intersections A number is divisible by both 4 and if and only if it is divisible by LCM(4, = 12 Use inclusion-exclusion principle Let A 1 be the set of all perfect squares from 1 to 10000, and A 2 be the set of perfect cubes Then A 1 = [ 10000] and A 2 = [10000 1/ ], A 1 A 2 = [10000 1/ ], where [x] is the largest integer that is less than or equal to x for any given real number 5 Do similarly as in Prob 4 Change into a question of combinations of multiset similar to problem 4 Or consider the set of all solutions with x i, at least 9 10 There is an r-combination means that the equation x 1 + x 2 + + x k = r has a non-negative integer solution with x 1 n 1, x 2 n 2,, x k n k In particular n 1 + n 2 + + n k r But the fact that A 1 A 2 A k is non-empty means that there integers x 1 > n 1, x 2 > n 2,, x k > n k such that x 1 + x 2 + + x k = r > n 1 + n 2 + n k 12 One has to consider which ( four integers are in their natural positions, while all other are not in their natural positions 4 ways to choose the four positions D4 ways to arrange the other four with none in its natural position Hence ( 4 D4 14 D4 See # 12 for argument ( 4 20 Note that D n = n(d n 1 + ( 1n n Then use induction on n 24(b The forbidden positions are divided into clusters r 1 = 4 r 2 = 2 + ( 2 4 4(The two positions are in the same cluster or in separate clusters r = P (, 22 4 + 4 4 4 (The three positions are in two clusters or in three clusters r 4 = ( 2 2 2 + ( 1 2 4 4 (Four positions are in two clusters or in three clusters with one of then have two positions r 5 = ( 1 4 2 2 (one of the cluster has two positions and the other two has two positions r = 2 2 2 2 Convert the problem into a chess-board with forbidden positions

27 Use circular permutations and inclusion-exclusion principle A 1 is all circular permutations with 12 appears, A 7 all circular permutations with 7 appears, and A all circular permutations with 1 appears 0 How many circular permutations are there of the multiset {2 a, b, 4 b, 5 d} where, for each type of letter, all letters of that type do not appear consecutively, (ie, not all letters of that type appear consecutively Solution Let S be the set of all circular permutations of the multiset Let A a be the set of all circular permutations with all a s appear consecutively Similarly, let A b, A c, and A d be the set of all circular permutations with all b s, c s, and d s, respectively, appear consecutively To compute A a, we change into the question of linear permutations with all a s appear at the very beginning Thus there are ( +4+5,4,5 many such linear permutations, ie, Aa = 12! Similarly,! 4! 5! A b = 11!, A 2! 4! 5! c = 10!, and A 2!! 5! a = 9! A 2!! 4! a A b = 10! (treating all a s as one a and 1! 4! 5! at the beginning to change into a linear permutation question and all b s as just one b Similarly, A a A c = 9!, A! 1! 5! a A d =!, A! 4! 1! b A c =!, A 2! 1! 5! b A d = 7!, A 2! 4! 1! c A d =!, 2!! 1! A a A b A c = 7!, A 1! 1! 5! a A b A d =!, A 1! 4! 1! a A c A d = 5!, A! 1! 1! b A c A d = 7!, 2! 1! 1! and A a A b A c A d =! 1! 1! 1! One needs to calculate S Note that since the greatest common divisor of 2,, 4, 5 is 1, each circular permutation of the multiset {2 a, b, 4 c, 5 d} corresponds to exactly 2 + + 4 + 5 different linear permutations of the multiset (why? Then S = (2++4+5! /(2 + + 4 + 5 Now 2!! 4! 5! use the inclusion and exclusion principle to compute Āa Āb Āc Ād Note that the circular permutation of { a, b, c} determined by the linear permutation abcbabcbabcb corresponds to only 4 different linear permutations Section 7 1 The characteristic equation is x x + 2 = 0 Observe that x = 1 is a solution and factorize to get (x 1 2 (x 2 Then general solution is h n = c 1 + c 2 n + c ( 2 n by Theorem 722 2(b n=0 ( 1n x n = 1 + 1 x 2(d 1 n=0 n! xn = e x 20(c g(x xg(x 9x 2 g(x + 9x g(x = h 0 + (h 1 h 0 x + (h 2 h 1 9h 0 x 2 = x + x 2 g(x = x + x 2 1 x 9x 2 9x = x + x 2 (1 + x(1 x(1 x = g(x = 1 1 x 2 1 1 x 5 1 1 x 1 1 x 7 1 Section 95 1 1 1 1 x + 12 1 + x + 1 x 7

R X X X R X X R X X X X X X R X X X R X X R 4 Any bipartite graph with at least one vertex meeting five or more edges will not be a domino-bipartite graph of any board 7 Count the number of edges through x-vertices and then through Y -vertices 9 Follow the matching algorithm starting a with matching (try to find the matching with as many edges as you can to save time 10 First select a matching M (with as many edges as you could find The follow the matching algorithm If you get a break-through, find the m-alternating chain and a new matching with more edges Then apply the matching algorithm with new matching and repeat till a nonbreak-through happens In that case the matching is a max matching 12 Check the Marriage Condition A 1 A A 4 A 5 = < 4 Having A 5 removed the remaining the sets has an SDR by checking the Marriage Condition 14 First take an SDR {e 1, e 2,, e n } for the collection A If x is one of the member, then it is an SDR we are looking for Otherwise, replace e 1 by x to get a new SDR {x, e 2,, e n } Example: A 1 = {a, x} and A 2 = {x} and A = (A 1, A 2 Then e 1 = a, e 2 = x give an SDR, but A 1 will never be represented by x as x has to represent A 2 1 A sequence of numbers e 1, e 2,, e n from {1, 2,, n} for an SDR for A with e i in A i if any only if that e i i and all are distinct Since there are only n different numbers, thus the sequence give forms a permutation of the set {1, 2,,n}, and thus a derangement Conversely, each derangement gives an SDR D n > 0 if n > 1 ( 1, 2 2, 1 19 2, 1 1, 2

21 Consider the preferential ranking matrix w 1 1, n 2, n 1, n 2 n, 1 w 2 n, 1 1, n 2, n 1 n 1, 2 w n 1, 2 n, 1 1, n n 2, w n 1, n 2 4, n 5, n 4 2, n 1 w n 2, n 1, n 2 4, n 1, n Prove that, for each k = 1, 2,, n, the complete marriage in which each woman gets her kth choice is stable Proof If each woman gets her kth choice, then w 1 m k, w 2 m k+1,, w n k+1 m n, w n k+2 m 1, w n w k 1 Note that m i is his wifes n k + 1th choice Note that for each pair (w i, m j the ranking p of m j by w i and the ranking q of w i by m j satisfy the equation p + q = n + 1 If w i ranks m j higher than her husband, then p < k then q > n k + 1 Thus m j ranks w i lower than his wife Thus the marriage is stable 9