Patel_Gunjan_Assignment___4

# Patel_Gunjan_Assignm - -Discrete Mathematics Assignment 4-|5.1 The Basics of Counting(page 346 Exercise 40| In how many ways can a photographer at

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Unformatted text preview: -----------------------------------------------------------------------------------------------------------------Discrete Mathematics Assignment # 4 -----------------------------------------------------------------------------------------------------------------|5.1 The Basics of Counting (page 346): Exercise 40| In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if (a) The bride must be in the picture? We know that one place (out of 6) are reserved for the bride. Therefore, we have 5 more places (6[Total] 1) left. There are, in total, 10 people. So there are 9 people when we take out the bride. So we need to CHOOSE 5 people from 9 people: 1 * 9 * 8 * 7 * 6 * 5 = 15120 Figure I Reserved for bride Since it is not specified where the bride is arranged, we cannot assume a specific location. Therefore, we must consider all 6 positions. The number of ways that 6 people can be arranged if bride is in first position (Figure I) is 15120. Since we have 6 positions, the final answer is 6 * 15120 = 90720. In other words: Ignoring the bride, we need to CHOOSE 5 people from a batch of 9 people. So we can conclude to: Since the question doesn't specify order we need to calculate To find the final answer we must multiply 126 with 720. 126*720 = 90720 -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------- (b) Both the bride and groom must be in the picture? We know that two places (out of 6) are reserved for the bride and groom. So ignoring the bride and groom, we need to CHOOSE 4 people from a batch of 8. We can do this by: Again the question doesn't specify the location of the bride and groom. Therefore we need to do: P(6,6) = 720. To find the total ways of arrangement, we must do: 70 * 720 = 50400. -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------(c) Exactly one of the bride and the groom is in the picture? We know that we can only CHOOSE ONE from the bride and groom. This means: After doing so, we must pick the remaining 5 people from a batch of 8. This means: Similarly to part (a) and (b), we are not given a specific order. Therefore, we must include P(6,6) = 720. The final answer is: 2 * 56 * 720 = 80640 -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------- |5.2 The Pigeonhole Principle (page 354): Exercise 30| Show that if there are 100,000,000 wage earners in the United States who earn less than 1,000,000 dollars, then there are two who earned exactly the same amount of money, to the penny, last year. Since the question says `wage earners' we can assume the first value of the sequence is \$0.01, because no one can earn \$0.00. The question also states that earners earn LESS than \$1,000,000. Therefore, we are looking for ALL values between \$0.00 and \$1,000,000 exclusive, which increase by \$0.01 each time because the question says `to the penny'. So the sequence goes like: \$0.01, \$0.02, \$0.03 ... \$999,999.99. In the sequence, there are 99,999,999 numbers. Since the amount of wage earners (100,000,000) is 1 more than the numbers in the sequence (99,999,999), there will be two people who earn exactly the same wage, to the penny. When referring to the pigeonhole principal, we can refer to wage earners as objects. We can also state that there are 99,999,999 boxes. Since there are more objects than there are boxes, we can conclude that two objects will be placed in the same box. --------------------------------------------------------------------------------------------------------------------- |5.3 Permutations and Combinations (page 362): Exercise 30| Seven women and nine men are on the faculty in the mathematics department at a school. (7 women + 9 men = 16 people [Total]) (a) How many ways are there to select a committee of five members of the department if at least one woman must be on the committee? This question states that there needs to be AT LEAST one woman. This is a restriction that we must consider. We will keep that in mind. Assuming that there are NO restrictions, we can CHOOSE 5 people for a batch of 16 people, in the following way: However, we have a restriction in play here. We need to have AT LEAST one woman in the committee. We know that are 9 men, and if we assign the committee with ALL men, we would do it in the following way: But we must include AT LEAST one women. We can accomplish that by doing the following: C(16,5) C(9,5) = 4368 126 = 4242 -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------- b) How many ways are there to select a committee of five members of the department if at least one woman and at least one man must be on the committee? In part (a), we needed to have at least one woman. This question is similar, but it adds one factor. We also need to have at least one man. So we can use the solution from part (a), and attach the AT LEAST one man part to it. C(16,5) C(9,5) - C (7,5) = 4368 126 - 21 = 4221 No Restriction All women in committee All men in committe e --------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------- |6.1 An Introduction to Discrete Probability (page 399): Exercise 14| What is the probability that a five-card poker hand contains cards of five different kinds? We know that there are 52 cards in a standard deck of cards. We also know that there are 13 unique different values (2,3,4,5,6,7,8,9,10,J,Q,K,A), that repeat 4 times, with 4 different suits (13 * 4 = 52 cards). We must CHOOSE 5 different cards for a batch of 13 different cards: For each card, we will CHOOSE a suit from a batch of 4 suits. Since we have 5 cards, we will need to repeat the process 5 times: There are a total of 52 cards, from which we need to select 5 cards. Therefore, the total possibilities we can have is: To calculate the final answer, we must do the following: In order to convert into percentage, we must do: 0.507 * 100% = 50.7% -------------------------------------------------------------------------------------------------------------------The probability that a five-card poker hand contains cards of five different kinds is 50.7%. --------------------------------------------------------------------------------------------------------------------- ...
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## This note was uploaded on 03/11/2012 for the course LSDHFA 234 taught by Professor Sdaf during the Spring '12 term at UOIT.

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