Loading in 5 sec....

Math 106 – Exam #3 - Review ProblemsPowerPoint Presentation

Math 106 – Exam #3 - Review Problems

- 62 Views
- Uploaded on

Math 106 – Exam #3 - Review Problems. 1. (a) (b) (c) (d).

Download Presentation
## PowerPoint Slideshow about '' - lavender

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Math 106 – Exam #3 - Review Problems

1.

(a)

(b)

(c)

(d)

A room contains 90 men where 25 are wearing a hat, 34 are wearing a coat, 33 are wearing a pair of boots, 8 are wearing a hat and a coat, 10 are wearing a hat and a pair of boots, 14 are wearing a coat and a pair of boots, and 3 are wearing a hat, a coat, and a pair of boots.

How many men in the room are wearing at least one of a hat, a coat, and a pair of boots?

25 + 34 + 33 – 8 – 10 – 14 + 3 = 63

How many men in the room are wearing no hat, no coat, and no boots?

90 – 63 = 27

How many men in the room are wearing a hat or a coat?

25 + 34 – 8 = 51

How many men in the room are wearing no coat and no boots?

90 – (34 + 33 – 14) = 90 – 53 = 37

(a)

(b)

(c)

(d)

Let the set of all different possible arrangements (permutations) of the letters AEIOUY (i.e., all the letters which can be vowels) be the universal set. We define the subsets

A = the set of all arrangements where the letters O and Y are together

B = the set of all arrangements where the letters A, E, and I are together

Find the size of the universe.

#U =

P(6,6) = 720

Find each of the following:

#A = #B =

#(AB) =

P(5,5) P(2,2) = 240

P(4,4) P(3,3) = 144

P(3,3) P(3,3) P(2,2)= 72

Find the number of arrangements where the letters O and Y are together and the letters A, E, and I are together.

#(AB) = 72

Find the number of arrangements where the letters O and Y are together or the letters A, E, and I are together.

#(AB) = #A + #B– #(AB) = 240 + 144 – 72 = 312

(a)

(b)

(c)

(d)

Let the set of all integers consisting of 5 digits selected from 1 to 9 with no repetition allowed be the universal set. We define the subsets

A = the set of integers where the digit 1 does not appear

B = the set of integers where the digits 8 and 9 do not appear

Find the size of the universe.

#U =

P(9,5) = 15,120

Find each of the following:

#A = #B =

#(AB) =

P(8,5) = 6720

P(7,5) = 2520

P(6,5) = 720

Find the number of integers where none of the digits 1, 8, and 9 appear.

#(AB) = 720

Find the number of integers where the digit 1 appears at least once.

#~A = #U– #A = 15,120 – 6720 = 8400

(a)

(b)

(c)

(d)

Let the set of all integers consisting of 5 digits selected from 1 to 9 with repetition allowed be the universal set. We define the subsets

A = the set of integers where the digit 1 does not appear

B = the set of integers where the digits 8 and 9 do not appear

Find the size of the universe.

#U =

95 = 59,049

Find each of the following:

#A = #B =

#(AB) =

85 = 32,768

75 = 16,807

65 = 7776

Find the number of integers where none of the digits 1, 8, and 9 appear.

#(AB) = 7776

Find the number of integers where the digit 1 appears at least once.

#~A = #U– #A = 59,049 – 32,768 = 26,281

(a)

(b)

Let the set of all integers from 1 to 5000 be the universal set. We define the subsets

A = the set of integers divisible by 9

B = the set of integers divisible by 21

C = the set of integers divisible by 33

Find the size of the universe.

#U =

5000

Find each of the following:

#A =

#B =

#C =

#(AB) =

555

19 = 9, 29 = 18, 39 = 27, …, 5559 = 4995, 5569 = 5004

238

121 = 21, 221 = 42, 321 = 63, …, 23821 = 4998, 23921 = 5019

151

133 = 33, 233 = 66, 333 = 99, …, 15133 = 4983, 15233 = 5016

79

9 = 33

21 = 37

Integers divisible by each of 9 and 21 must be divisible by 337 = 63

163 = 63, 263 = 126, 363 = 189, …, 7963 = 4977, 8063 = 5040

#(AC) =

#(BC) =

#(ABC) =

50

9 = 33

33 = 311

Integers divisible by each of 9 and 33 must be divisible by 3311 = 99

199 = 99, 299 = 198, 399 = 297, …, 5099 = 4950, 5199 = 5049

21

21 = 37

33 = 311

Integers divisible by each of 21 and 33 must be divisible by 3711 = 231

1231 = 231, 2231 = 462, 3231 = 693, …, 21231 = 4851, 22231 = 5082

7

9 = 33

21 = 37

33 = 311

Integers divisible by each of 9, 21, and 33 must be divisible by 33711 = 693

1693 = 693, 2693 = 1386, 3693 = 2079, …, 7693 = 4851, 8693 = 5544

(c)

(d)

(e)

Find the number of integers which are divisible by all of 9, 21, and 33.

#(ABC) = 7

Find the number of integers which are divisible by at least one of 9, 21, and 33.

#(ABC) = 555 + 238 + 151 – 79 – 50 – 21 + 7 = 801

Find the number of integers which are divisible by none of 9, 21, and 33.

#~(ABC) = 5000 – 801 = 4199

(a)

(b)

Ten identical candy bars (all Snicker’s) are available for second-grade teacher Rachael to distribute to the other teachers in the lounge. The other teachers in the lounge are named Casey, Dan, Kathy, and Mary.

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself?

c + d + k + m + r = 10

non-negative integers

14!

——— = 1001

4! 10!

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if she does not keep more than 3 for herself?

c + d + k + m + r = 10

r 3

Use GOOD = ALL – BAD

10!

——— = 210

4! 6!

c + d + k + m + r = 10

4 r

1001 – 210 = 791

(d)

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if Kathy and Mary each do not want more than 2?

c + d + k + m + r = 10

k 2 AND m 2

Use GOOD = ALL – BAD

c + d + k + m + r = 10

3 k OR 3 m

1001 – 590 = 411

11! 11! 8!

——— + ——— – ——— = 330 + 330 – 70 = 590

4! 7! 4! 7! 4! 4!

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if Casey does not want more than 5, and Dan does not want more than 4?

c + d + k + m + r = 10

c 5 AND d 4

Use GOOD = ALL – BAD

c + d + k + m + r = 10

6 k OR 5 m

1001 – 196 = 805

8! 9!

——— + ——— – 0 = 70 + 126 – 0 = 196

4! 4! 4! 5!

6.-continued

(e)

(f)

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if no one gets more than 3?

Use GOOD = ALL – BAD

c + d + k + m + r = 10

c 3 AND d 3 AND k 3 AND m 3 AND r 3

c + d + k + m + r = 10

4 c OR 4 d OR 4 k OR 4 m OR 4 r

10! 6!

(5)—— – (10)—— + 0 = 900

4! 6! 4! 2!

1001 – 900 = 101

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if no one gets more than 2?

one (1), since c = 2, d = 2, k = 2, m = 2, r = 2 is the only possible solution

How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself, if she allows herself only one and no one else gets more than 2?

zero (0), since this is impossible

Download Presentation

Connecting to Server..