#$&* course 152 9/14 2pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution:: ** There are 25 students available so there are 25 choices for the first student. On the second choice there are 24 students left so there would be 24 possibilities. Similarly on the third, fourth and fifth selections there would be 23, 22 and 21 choices, respectively. The result, by the Fundamental Counting Principle, is 25 * 24 * 23 * 22 * 21 choices. 25 * 24 * 23 * 22 * 21 = 25 ! / ( (25 - 5) !) since 25 ! / ( (25 - 5) !) = 25 ! / (20! * 5!) = 25 * 24 * 23 * 22 * 21 25 ! / ( (25 - 5) !) is P(25, 5). We use permutations because in this case, there are 5 different prizes so the order in which the students are chosen makes a difference in the final outcome. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My main problem on this one was deciding which form to use for my final answer ********************************************* question: Is repetition allowed in this situation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: No because there is only one winner for each prize and each winner can only win once confidence rating #$&* 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 #$&* question 11.3.30 3-letter monogram all letters different YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1. First we have to find out how many selections we have and we know that the first letter cannot be Y or Z 2. Now we have to find out how many choices we have for the second we know it that it is dependent on the first letter because it must be in Alphabetical order and it is dependent on the third so it can’t be Z 3. And we know that the third letter is going to be Z 24*24*1 / 2 = 288 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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 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): I don’t believe that you have this one correct you say the there are 25 choices for the first but that’s not possible because if you have to be in Alphabetical order then you wouldn’t be able to use Y for the simple fact there must be a middle letter that follows Y before Z ------------------------------------------------ Self-critique rating #$&* question 11.3.43 /& 42 divide 25 students into groups of 3,4,5,6,7. In how many ways can the students be grouped? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For the group of three you would get C(25,3) = 25! ((25-3)!) = 25*24*23 For the group of four you would get C(22,4)=22!((22-4)!) = 22*21*20*19 For the group of five you would get C(18,5)=18!((18,5)!) = 18*17*16*15*14 For the group of six you would get C(13,6)=13!((13-6)!) = 13*12*11*10*9*8 For the group of seven you would get P(7,7) = 7!((7-7)!) = 7*6*5*4*3*2*1 25! / (3!*4!*5!*6!*7!) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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 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. 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 have exactly 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 #$&* question 11.3.60. C(n,0)What is the value of C(n,0)? What is the value of C(8,0)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: C(n,0) = zero because there is nothing to choose from so no matter what n = the answer will always be 0 C(8,0) = zero like I just explained there is no options to pick from
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Given Solution: ** C(n, r) is the number of ways of choosing r objects out of n available objects, without regard for order. C(n, 0) is therefore the number of ways to choose 0 objects from among n objects. No matter what n is, there is exactly one way to do this, which is to choose nothing. Thus C(n, 0) is always equal to zero. As another example: There are C(4,2) = 6 ways in which to obtain 2 Heads on four flips of a coin, C(4,3) = 4 ways to obtain 3 Heads, C(4,4) = 1 way to obtain 4 Heads. Obtaining 4 Heads is the same as obtaining 0 Tails, and of course C(4,0) is the number of ways to obtain 0 Tails. So C(4,0) must be 1. The formula also gives us the same result: C(n, 0) = n ! / [ (n - 0) ! * 0 ! ] = n ! / ( n ! * 1) = 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 004. `query 4 question 11.4.6 Find C(9,6) on Pascal's triangle. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 9! / (6! (6-3)!) = 9*8*7/(3*2*1) = 3*4*7 = 84 Go down to the row that says 9 and over to the 6 position but you have to remember to skip the 0 position confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution:: ** C(9, 6) occurs in the n=9 row, the r=6 position of Pascal’s Triangle, which is the 10th row and the 7th number in the row. C(9,6) = 9! / ( 6! (6-3)! ) = 9*8*7 / (3*2*1) = 3 * 4 * 7 = 84 does agree with the number in the n = 9 row and the r = 6 position. (Note that since the first row is row 0 and the first element in every row is element 0, the n = 9 row is the 10th row, and the r = 6 position is the 7th number from the left.) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.4.18 clueless check of four of nine possible classrooms. How many of the possible selections will fail to locate the classroom? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There are C(9,4) = 9!/(4!(9-4)!) There are 126 different combinations And 8 out of the 9 are wrong C(8,4) = 8!/(4!(8-4)!) 70 out of the 126 will be wrong confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** There are C(9,4) possible combinations of the four classrooms to be checked. There are 8 'incorrect' classrooms, so there are C(8, 4) ways for the check of 4 rooms to yield a 'wrong' classroom. C(9,4) = 126 and C(8,4) = 70. So there are 126 possible combinations of the four classrooms to be checked, and 70 of these combinations fail to include the ‘right’ classroom. Thus only 126 – 70 = 56 of the possible combinations will include the ‘right’ classroom. So the chance of finding the right classroom is 56 / 126, a little less than 50%. STUDENT QUESTION Can you work out how C (9,4) = 126 and how C (8,4) = 70??? Where did I go wrong??? I worked them out the way I thought I should. Am I getting combinations confused with permutations, or do I just have it totally wrong??? INSTRUCTOR RESPONSE I think you have mixed up permutations and combinations. You calculuated 9! / (9-4)!.. To clarify: C(9, 4) = 9! / ( 4! * (9-4)! ), not 9! / (9-4)!. 9! / (9-4)! = 9 * 8 * 7 * 6, which is the number of possible ordered choices of 4 objects chosen from among 9. This is the permutation. To get the number of unordered choices, you would divide the number of ordered choices by 4! (since every set of 4 choices could have appeared in any of 4! orders). So instead of 9! / (9-4)!., you would divide this by 4! to get 9! / ( 4! (9-4)! ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.4.31 /& 30 What sequence by totaling diagonals of Pascal's Triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1,1,2,3,5,8,13,21,34,55 Each number in the pattern is equal to the two previous numbers add together And I believe the pattern is called the Fibonacci Sequence which states the first two numbers are 1 , 1 and then rest of the pattern is laid out by adding yhe two preceding numbers confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** The totals are 1 1 1 + 1 = 2 1 + 2 = 3 1 + 3 + 1 = 5 Etc. It is easy to see that the numbers are 1, 1, 2, 3, 5, 8, 13, 21, ... and that these numbers form what is called the Fibonacci Sequence, which seems to occur in all sorts of unexpected places. The rule for the Fibonacci sequence is that the first two numbers are both 1. Then each subsequent number is the sum of the two numbers that precede it.** Regrettably, subsequent problems in this section have been omitted from Section 11.4, as of the 11th edition of the text. The remainder of this Query may therefore be omitted. Query 11.4.42 xxxxxxxxxx (x+y)^8 **(x + y ) ^ 8 = x^8 + C(8,1) x^7 y + C(8,2) x^6 y^2 + C(8,3) x^5 y^3 + C(8,4) x^4 y^4 + C(8,5) x^3 y^5 + C(8,6) x^2 y^6 + C(8,7) x^7 y + y^8 = x^8 + 8 x^7 y + 28 x^6 y^2 + 56 x^5 y^3 + 70 x^4 y^4 + 56 x^3 y^5 + 28 x^2 y^6 + 8 x y^7 + y^8. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. ** STUDENT COMMENT: I was ok with this assignment until I got to problem 11.4 - 42 I do not understand the reasoning behind the following problems. 11.4 - 45: (2a + 5b)^4 = The binomial expansion is listed in the answer section, but I do not understand how they got there. INSTRUCTOR RESPONSE: Here is the solution for (2a + 5b) ^ 5. The application of the Binomial Theorem is clearer for the 5th power than the 4th; if you understand this you'll get the pattern for the 4th power. The answer is found from C(5,5) * (2a)^2 * (5b)^0 + C(5, 4) * (2a)^4 * (5b)^1 + C(5, 2) * (2a)^3 * (5b)^2 + . etc., following the pattern of the binomial expansion formula. Expanding the powers of 2a and 5b we get C(5, 5) * 32 a^5 + C(5, 4) * 16 a^4 * 5b + C(5, 3) * 8 a^3 * 25 b^2 + etc., which is equal to 1 * 32 a^5 + 5 * 16 a^4 * 5b + 10 * 8 a^3 * 25 b^2 + etc., or finally to 32 a^5 + 80 a^4 + 2000 a^3 + etc.. ANOTHER QUESTION: 11.4 -50 and 51 xxxxxxxxx The rth or general term of the binomial expansion for (x = y)^n and (x + y)^14;5th term. INSTRUCTOR RESPONSE: The rth term of (x+y)^n will be C(n, r) * x^r * y^(n-r). You are choosing x from r of the binomials and y from the remaining n - r binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) . (x+y), where it is understood that we have (x+y) written n times. The 5th term of (x+y)^14 requires that you choose x from 5 of the binomials and y from the other 14-5 = 9 binomials in the expression (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y). There are C(14, 5) ways to do this, and the result for every one of these ways is x^5 * y^14. So the 5th term is C(14,5) x^5 y^9. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "