challset1sol

# Pigeon hole principle one of months holes will have

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pigeon hole principle, one of months (holes) will have two people who have a birthday in it. b. Since the numbers {1,2,…,10} are either even or odd, it is necessary to break this down into cases. case 1. 5 even integers, and 1 odd integer. Since we picked all the even integers, 2 divides 4, 6,8,10. We have a number that divides another exactly. case 2. 4 even integers, and 2 odd integers. If 2 and 1 are selected then we are done. If we don’t have 2 , then the four even integers left are 4,6,8,10, and since 4 divides 8 exactly, we have a number which divides another exactly. case 3. 2 even integers and 4 odd integers. If 2 and 1 are selected then we are done (i.e. we have an integer which divides another one exactly). If we don’t have 1 , then the four odd integers left are 3,5,7,9, and since 3 divides 9 exactly, we have a number which divides another exactly. case 4. 1 even integer and 5 odd integers. Since all the odd integers have been picked, the number 1 has been picked and 1 divides every integer exactly. case 5. 3 even integers and 3 odd integers. This is the most difficult case. If 2 and 1 are selected then we are done. We are now working with the four even integers, 4,6,8,10 and the four odd integers, 3,5,7,9. There are four ways to select three integers from a set of four integers. We could examine the 4x4=16

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cases that arise, and discover that in each case, there will be one integer that divides another exactly, however with some selectivity, we can shorten this process. Since 4 divides 8 and 3 divides 9, it follows that we don’t need to have 4 and 8, or 3 and 9 together when we select three even and three odd integers. If they are together then there is an integer which divides another one exactly. The following are the remaining cases: i. {4,6,10}, {3,5,7}- 3 divides 6 exactly. ii. {4,6,10}, {5,7,9}- 5 divides 10 exactly. iii. {6,8,10}, {3,5,9}- 3 divides 6 exactly. iv. {6,8,10},{5,7,9}- 5 divides 10 exactly. 5. Let teams A,B,C, and D play in a round robin tournament. The following chart gives an example of what could happen. A B C D Points A -- W W W 3 B L -- L W 1 C L W -- W 2 D L L L -- 0 Total 6 With four teams in a round robin tournament there will be 6 games played. As a result of this there will be 6 wins and 6 losses arising. The six wins will account for a total of 6 points. Hence the sum of the scores doesn’t depend upon the outcome of the games. 6. For any arrangement of men on the dance floor, there are 5x4x3x2x1=5!=120 ways the women can be matched with the men. Another approach to this is to pick the 5 couples first, i.e.
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• Fall '06
• miller
• Natural number, Prime number, positive integer

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