#$&* course Mth 152 1/29 8:49 Your solution, attempt at solution.
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Given Solution:: ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices, respectively. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 25 ! / ( (25 - 5) !) is P(25, 5). We use permutations because in this case, there are 5 different prizes so the order in which the students are chosen makes a difference in the final outcome. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* question: Is repetition allowed in this situation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: No, repetition is not allowed for permutations by definition. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** GOOD STUDENT ANSWER: no repetition is allowed because there are 5 different prizes, and you can't give the same one to two people ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: question 11.3.30 3-letter monogram all letters different YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We use combinations so C(25,2) = 25!/2!23! = 300 ways. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** We are choosing 3 different letters, and since the monogram will be different if you change the order of the letters, we can say thatorder 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 possible objects. P(26,3) = 26!/(26-3) ! = 26 * 25 * 24, in agreement with the previous expression. However in this question 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. ** MODIFIED SOLUTION: I believe in the original solution that I overlooked the requirement that the letters be in alphabetical order. Z is the last letter, so as long as the other two are chosen from the first 25 letters of the alphabet, it will be possible to construct the monogram. Any combination of two of the 25 remaining letters can be used. Once the combination is selected, the letters will then be put into alphabetical order. There are C(25, 2) = 25 * 24 / 2! = 300 possible combinations, so there are 300 possible monograms with the letters distinct and fin order. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: question 11.3.43 /& 42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We use C(25,3) and C(22,4) and C(18,5), and C(13,6) and finally the last group is C(7,7) so we then use FCP and multiply the groups together. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** We can choose the groups in any order we wish. Each group chosen is chosen without regard for order. If we choose to begin by making the group of 3, there are 25 students available when we begin to select our group, so 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 says 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 thenumber 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. So get have • C(25,3) * C(22,4) * C(18,5) * C(13,6) * C(7,7) ways to complete the grouping. 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 complete the groupings. The question is, would this make a difference in the final result? To find out we compare the two results 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). 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, both turn out to haveexactly the same numbers in their numerators, and the same numbers in their denominators. As a result, they both end up in the same form o 25! / [ 3 ! * 4 ! * 5 ! * 6 ! * 7 ! ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: question 11.3.60. C(n,0)What is the value of C(n,0)? What is the value of C(8,0)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: C(N,0) is the number of ways to chose zero objects from among N objects. It is equal to zero.
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Given Solution: ** C(n, r) is the number of ways of choosing r objects out of n available objects, without regard for order. C(n, 0) is therefore the number of ways to choose 0 objects from among n objects. No matter what n is, there is exactly one way to do this, which is to choose nothing. Thus C(n, 0) is always equal to zero. 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) is the number of ways to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us the same result: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " self-critique #$&* #$&* self-critique self-critique rating " self-critique #$&* #$&* self-critique self-critique rating #(*!