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Math 106 – Exam #3 - Review Problems

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

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Math 106 – Exam #3 - Review Problems

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  1. 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

  2. 2. (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 = #(AB) = 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. #(AB) = 72 Find the number of arrangements where the letters O and Y are together or the letters A, E, and I are together. #(AB) = #A + #B– #(AB) = 240 + 144 – 72 = 312

  3. 3. (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 = #(AB) = 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. #(AB) = 720 Find the number of integers where the digit 1 appears at least once. #~A = #U– #A = 15,120 – 6720 = 8400

  4. 4. (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 = #(AB) = 85 = 32,768 75 = 16,807 65 = 7776 Find the number of integers where none of the digits 1, 8, and 9 appear. #(AB) = 7776 Find the number of integers where the digit 1 appears at least once. #~A = #U– #A = 59,049 – 32,768 = 26,281

  5. 5. (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 = #(AB) = 555 19 = 9, 29 = 18, 39 = 27, …, 5559 = 4995, 5569 = 5004 238 121 = 21, 221 = 42, 321 = 63, …, 23821 = 4998, 23921 = 5019 151 133 = 33, 233 = 66, 333 = 99, …, 15133 = 4983, 15233 = 5016 79 9 = 33 21 = 37 Integers divisible by each of 9 and 21 must be divisible by 337 = 63 163 = 63, 263 = 126, 363 = 189, …, 7963 = 4977, 8063 = 5040

  6. #(AC) = #(BC) = #(ABC) = 50 9 = 33 33 = 311 Integers divisible by each of 9 and 33 must be divisible by 3311 = 99 199 = 99, 299 = 198, 399 = 297, …, 5099 = 4950, 5199 = 5049 21 21 = 37 33 = 311 Integers divisible by each of 21 and 33 must be divisible by 3711 = 231 1231 = 231, 2231 = 462, 3231 = 693, …, 21231 = 4851, 22231 = 5082 7 9 = 33 21 = 37 33 = 311 Integers divisible by each of 9, 21, and 33 must be divisible by 33711 = 693 1693 = 693, 2693 = 1386, 3693 = 2079, …, 7693 = 4851, 8693 = 5544 (c) (d) (e) Find the number of integers which are divisible by all of 9, 21, and 33. #(ABC) = 7 Find the number of integers which are divisible by at least one of 9, 21, and 33. #(ABC) = 555 + 238 + 151 – 79 – 50 – 21 + 7 = 801 Find the number of integers which are divisible by none of 9, 21, and 33. #~(ABC) = 5000 – 801 = 4199

  7. 6. (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

  8. (c) (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!

  9. 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

  10. (g) 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

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