组合数学第三版 答案
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Spring 2007 Math 510 Hints for practice problems
Section 1.8. 3. Imagine a prison consisting of 64 cells arranged like the squares of an 8 × 8-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?) 13. Read the algorithm of de la Loub` ere’s method in the book. Section 2.4. 2. Proof. Suppose that 100 numbers are chosen. We write each number a = 2k a0 with a0 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 ofHale Waihona Puke Baiduthe 100 selected numbers. Let a be one of 100 selected number such that a ≤ 15. Write a = 2k a0 with a0 odd. Thus 0 ≤ k ≤ 3. If k = 0, then a = a0 is odd and, 3a, 5a, 7a, 9a, 11a, 13a are odd numbers no larger than 199. So they are odd parts of some selected numbers. Hence, a divides at least 6 selected numbers. If 2 ≥ k > 0, then a0 < 8. Thus, a0 ≤ 7. In this case, 3a0 , 9a0 and 27a0 are odd and less than 200 (27 × 7 = 189), thus they are the odd parts of some of the 100 selected numbers, say that they are b1 = 2e1 3a0 , b2 = 2e2 9a0 , and b3 = 2e3 27a0 . If one of e1 , e2 , e3 is at least 2, then a divides one of numbers b1 , b2 b3 . Otherwise two of e1 , e2 , e3 equal by the Pigeonhole principle. Then two of b1 , b2 , b3 with equal powers of 2 will have one divide the other. If k = 3, then a0 < 2, thus a0 = 1 and 3a0 , 9a0 , 27a0 , 81a0 are odd and less than 200, thus are odd parts of some of the 100 selected numbers, say that they are b1 = 2e1 3a0 , b2 = 2e2 9a0 , b3 = 2e3 27a0 , and b4 = 2e4 81a0 . If one of the four powers of 2 is at least 3, then a divides the corresponding number. Otherwise, 0 ≤ ei ≤ 2 for i = 1, 2, 3, 4. Pigeonhole principle implies that two of the powers of 2 in the four numbers b1 , b2 , b3 , b4 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 2e3 3a0 , 2e5 5a0 , 2e7 7a0 , 2e9 9a0 , 2e11 11a0 , 2e13 13a0 are in the selection in addition to a for some e3 , e5 , e7 , e9 , e11 , e13 . 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. 3. 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 210 = 1024 different groups, while the group ages have to maximal of 10 × 60 = 600. 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 a1 , a2 , . . . , an , an+1 , there exist two of the integers ai and aj with i = j such that ai − aj is divisible by n. Proof. Hint: Consider the n boxes labeled by 0, 1, . . . , n − 1. Place the integers ai in the box labeled r is the remainder of ai is r when divided by n. Pigeonhole principle implies that there is a box with two difference objects ai and gj with i j , i.e., ai and aj have the same remainder when divided by n. Hence the difference of these two numbers have the remainder 0 when divided, i.e., ai − aj is divisible by n. 17. Write ai = 2mi to be the number of acquaintances of the ith person. Then 0 ≤ mi ≤ 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 98 people not including him/herself. That is impossible. 19. (a). Divide the triangle into four (1 + 3) equilateral triangles with sides length 1/2. (b) Divide the triangle into nine (1 + 3 + 5) equilateral triangles with sides length 1/3. (c) Divide the triangle into (1 + 3 + 5 + · · · + (2n − 1)) equilateral triangles with sides length 1/n. Section 3.6 1. easy. 5. Count the number of the factors 2 and 5. 8. 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 10 Among them, the number of committees without any women is 5 and the number of the committee with exactly one woman is 12 · 10 . Thus the number of committees with at least two 4 22 10 10 women is 5 − 5 − 12 · 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 3 9 − 9 restrictions. But the number of such committees with exactly one woman is 3 . Thus 20 3 3 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 − 10 − 12 · 10 − ( 20 − 9 ). 5 5 4 3 3 2
Section 1.8. 3. Imagine a prison consisting of 64 cells arranged like the squares of an 8 × 8-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?) 13. Read the algorithm of de la Loub` ere’s method in the book. Section 2.4. 2. Proof. Suppose that 100 numbers are chosen. We write each number a = 2k a0 with a0 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 ofHale Waihona Puke Baiduthe 100 selected numbers. Let a be one of 100 selected number such that a ≤ 15. Write a = 2k a0 with a0 odd. Thus 0 ≤ k ≤ 3. If k = 0, then a = a0 is odd and, 3a, 5a, 7a, 9a, 11a, 13a are odd numbers no larger than 199. So they are odd parts of some selected numbers. Hence, a divides at least 6 selected numbers. If 2 ≥ k > 0, then a0 < 8. Thus, a0 ≤ 7. In this case, 3a0 , 9a0 and 27a0 are odd and less than 200 (27 × 7 = 189), thus they are the odd parts of some of the 100 selected numbers, say that they are b1 = 2e1 3a0 , b2 = 2e2 9a0 , and b3 = 2e3 27a0 . If one of e1 , e2 , e3 is at least 2, then a divides one of numbers b1 , b2 b3 . Otherwise two of e1 , e2 , e3 equal by the Pigeonhole principle. Then two of b1 , b2 , b3 with equal powers of 2 will have one divide the other. If k = 3, then a0 < 2, thus a0 = 1 and 3a0 , 9a0 , 27a0 , 81a0 are odd and less than 200, thus are odd parts of some of the 100 selected numbers, say that they are b1 = 2e1 3a0 , b2 = 2e2 9a0 , b3 = 2e3 27a0 , and b4 = 2e4 81a0 . If one of the four powers of 2 is at least 3, then a divides the corresponding number. Otherwise, 0 ≤ ei ≤ 2 for i = 1, 2, 3, 4. Pigeonhole principle implies that two of the powers of 2 in the four numbers b1 , b2 , b3 , b4 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 2e3 3a0 , 2e5 5a0 , 2e7 7a0 , 2e9 9a0 , 2e11 11a0 , 2e13 13a0 are in the selection in addition to a for some e3 , e5 , e7 , e9 , e11 , e13 . 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. 3. 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 210 = 1024 different groups, while the group ages have to maximal of 10 × 60 = 600. 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 a1 , a2 , . . . , an , an+1 , there exist two of the integers ai and aj with i = j such that ai − aj is divisible by n. Proof. Hint: Consider the n boxes labeled by 0, 1, . . . , n − 1. Place the integers ai in the box labeled r is the remainder of ai is r when divided by n. Pigeonhole principle implies that there is a box with two difference objects ai and gj with i j , i.e., ai and aj have the same remainder when divided by n. Hence the difference of these two numbers have the remainder 0 when divided, i.e., ai − aj is divisible by n. 17. Write ai = 2mi to be the number of acquaintances of the ith person. Then 0 ≤ mi ≤ 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 98 people not including him/herself. That is impossible. 19. (a). Divide the triangle into four (1 + 3) equilateral triangles with sides length 1/2. (b) Divide the triangle into nine (1 + 3 + 5) equilateral triangles with sides length 1/3. (c) Divide the triangle into (1 + 3 + 5 + · · · + (2n − 1)) equilateral triangles with sides length 1/n. Section 3.6 1. easy. 5. Count the number of the factors 2 and 5. 8. 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 10 Among them, the number of committees without any women is 5 and the number of the committee with exactly one woman is 12 · 10 . Thus the number of committees with at least two 4 22 10 10 women is 5 − 5 − 12 · 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 3 9 − 9 restrictions. But the number of such committees with exactly one woman is 3 . Thus 20 3 3 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 − 10 − 12 · 10 − ( 20 − 9 ). 5 5 4 3 3 2