Math Module 4 Homework PDF

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1) Suppose that a company will select 3 people from a collection of 15 applicants to serve as a regional manager, a branch manager, and an assistant to the branch manager. In how many ways can the selection be made? Explain how you got your answer. P =

15 3

15 12

= 15 ∙ 14 ∙ 13 = 2,730

2) How many distinguishable permutations can be made of the letters in the word POSSIBILITIES? Explain how you got your answer. In the word POSSIBILITIES there are only two repetitive letters S and I, containing 3 S’s and 4 13 ! which is calculated to be 43,243,200. I’s. The equation will look something like: 4!3! 3) There are two display cases, one with 18 flowers and one with 15 flowers. How many ways can you choose 3 flowers from the first case and 2 from the second case? I use the equations C ∙ r! = nPr =

n! ( n−r ) !

and C =

n! r ! ( n−r ) !

Task 1 is to choose 3 flowers from the case with 18 flowers in it & Task 2 is to choose 2 flowers from the case with 15 flowers in it 18 ! . After 3 ! 15 ! calculating, this turns out to be 816 ways to choose 3 flowers from a case with 18 flowers in it. In the first task the equation will look something like this: 18C3 which equals

15! . 2 ! 13 ! After calculating, this turns out to be 105 different ways to choose 3 flowers from a case with 18 flowers in it. In the second task the equation will look something like this:

C which equals

15 2

In order to find the number of ways there are to choose 3 flowers from the first case and 2 from the second, you would multiply the number of ways there are from each task, which would be 816 ∙ 105 which equals 85,680.

4) A fair 6-sided die is rolled 5 times and the result is recorded for each roll. a) How many different sequences of results are possible? Explain how you got your answer. Since the dice has 6 sides and is being rolled 5 times, the number of different sequences that are possible are 65 or 7776 possible different sequences.

b) Of the possible sequences of results, how many of them contain exactly 3 rolls of a 4? Explain how you got your answer. Since we are looking for how many possible sequences contain exactly 3 rolls of a 4, 5 , which equals 10. the first thing I would do to my equation is: 3

()

I would then take the first part of the equation and multiply it by 5 2, because after the 3 rolls that will contain a 4, there are still 2 rolls left in the sequence. 5 ! 52 3!2! which would contain 3 rolls of a 4.

The equation would look like:

which can be calculated out to be 250 rolls

5) Show that if 17 integers from 1 to 32 are chosen, then there will be 2 of them that add up to 33. A1 = [1,32] A2 = [2,31] A3 = [3,30] A4 = [4,29] A5 = [5,28] A6 = [6,27] A7 = [7,26] A8 = [8,25] A9 = [9,24] A10 = [10,23] A11 = [11,22] A12 = [12,21] A13 = [13,20] A14 = [14,19] A15 = [15,18] A16 = [16,17] Considering that there 16 sets of integers listen, it can be true that any 2 of 17 integers from 1 to 32 will equal 33. 6) Show that there must be at least 59 ways to choose 5 integers from 1 to 15 so that all the choices have the same sum. Since there are 5 integers to be chosen from 1 to 15 the equation would look something like 15 which equals 3,003. 5 The minimum sum would be 1 + 2 + 3 + 4 + 5 which equals 15.

( )

While the maximum sum would be 11 + 12 + 13 + 14 + 15, which equals 65. This is followed by the equation 65 – 15 + 1 which equals 51. Finally, I take the first number found (3,003) and divide it by the number I found from the last equation (51), which ends up being 58.88 which could also be thought of as ≈ 59. 7) Jack is climbing stairs, taking one or two steps at a time. Let s n be the number of ways that Jack can climb n steps. a) Give a recurrence relation for s n. Explain your equation and be sure to include the initial conditions. S1 = 2 S2 = 3 Sn = Sn-1 + Sn-2 Since Sn is the number of ways jack can climb the steps, this equation demonstrates the ways he can climb depending on the number of stairs he climbs. b) In how many ways can Jack climb 10 steps? The number of ways Jack can climb 10 steps is represented with the equation: S 10 = S10-1 + S10-2 which equals S9 + S8. After following the sequence up to 10: S1 = 1 S2 = S1 + 1 = 2 S3 = S1 + S2 = 3 S4 = S2 + S3 = 5 S5 = S3 + S4 = 8 S6 = S4 + S5 = 13 S7 = S5 + S6 = 21 S8 = S6 + S7 = 34 S9 = S7 + S8 = 55 S10 = S8 + S9 = 89 Therefor, there are 89 way to climb 10 steps either 1 or 2 at a time. 8) Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. This equation can also be written as: x2 = -3x + 10 Which can be further evaluated as x2 + 3x – 10 = 0. The first five terms of the recurrence relation are 29, -47, -199, -503, -1111

b) Solve this recurrence relation. Show your reasoning. Two additional equations I have found useful are a 1 = u(1) + v(2) and a2 = u(1)2 + v(2)2. If a1 = 29 and a2 = -47, then: 29 = u(1) + v(2) and -47 = u(1) 2 + v(2)2. In solving the recurrence relation, I have found u to be 105 and v to be -38. 29 = 105(1) + -38(2) = 105 – 76 = 29 -47 = 105(1)2 + -38(2)2 = 105 – 152 = -47 c) Using the explicit formula you found in part b, evaluate a 5. You must show that you are using the equation from part b. 105(1)5 + -38(2)5 = ? 105 -1216 = -1111...