course Mth 152 ꮮʏ䞤x}d^assignment #003
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20:43:03 Query 11.3.20 5 prizes among 25 students
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RESPONSE --> 25 P 5 = 25!/ (25-5)! = 25!/ 20! = 6375600
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20:44:26 ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 This is P(25, 5). We use permutations because order matters when there are 5 different prizes. **
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RESPONSE --> I understand that I worked mine all of the way out, getting a solution with the same meaning as this
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20:45:02 Is repetition allowed in this situation?
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RESPONSE --> no
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20:45:54 ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people **
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RESPONSE --> I understand that you use permentation because you cannont give the same prize to 2 children
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20:47:16 Query 11.3.30 3-letter monogram all letters different
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RESPONSE --> 26 P 3= 26!/ (26-3)! = 26!/ 23! = 15600
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20:49:20 ** You are choosing 3 different letters, and since the monogram will be different if you change the order we can say that order definitely applies. If there is no restriction on any letter, other than the restriction of no repetitions, then there are 26 choices for the first letter, 25 for the second, 24 for the third so by the Fundamental Counting Principle there are 26 * 25 * 24 ordered choices. We can write this as P(26, 3), the number of possible permutations of 3 objects chosen without replacement from 26. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However the third initial must be the same as Judy's, which is `z'. Thus, since there can be no repetitions, there are only 25 possibilities for the first letter (can't be `z') and 24 for the second (can't be `z', can't be the first). So there are only 25 * 24 = 600 possibilities. **
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RESPONSE --> I understand that I gave the correct solution
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20:54:35 Query 11.3.42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped?
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RESPONSE --> 25 C 5 = 53130
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20:58:29 ** If we make the group of 3 first there are C(25, 3) possible choices. If we make the group of 4 next there are 22 students left from whom to choose so there are C(22, 4) possible choices. If we make the group of 5 next there are 18 students left from whom to choose so there are C(18, 5) possible choices. If we make the group of 6 next there are 13 students left from whom to choose so there are C(13, 6) possible choices. If we make the group of 7 next there are 7 students left from whom to choose so there are C(7, 7) possible choices. The Fundamental Counting Principle tells you that you have to multiply the number of ways of obtaining the first group by the number of ways of obtaining the second group by the number of ways of obtaining the 3rd group by the number of ways of obtaining the fourth group by the number of ways of obtaining the fifth group: C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7). Note that we could have chosen the groups in a different order, perhaps with the group of 7 first, the group of 6 second, etc.. The same reasoning would tell us that there are now C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) ways to do this. If the two expressions C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) and C(25, 7) * C(18, 6) * C(12,5) * C(7, 4) * C(3, 3) are both written down and simplified can get them both into the same form 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ], which can then be further simplified by cancellation and then multiplied. **
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RESPONSE --> I now understand hiw to answer that question. I understand that everytime you make a group you must subtract that number of people from the possibilites for the 2nd group. Thus, 25 C 3 * 22 C 4 *18 C 5 * 13 C6 * 7 C 7
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21:03:56 Query 11.3.60 C(n,0)What is the value of C(n,0)?What is the value of C(8,0)?
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RESPONSE --> C (2,0) = 2!/ O! (2-0)!= 2!/ 0!* 2!= 2/2
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21:06:13 ** This is equivalent to choosing 0 objects from n objects. No matter what n is, there is one way to do this, which is to choose nothing. As another example there are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) tells you how many ways that are to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us this: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. **
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RESPONSE --> I understand that 0! =1 therefore, anything nC0 will be 1.
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