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Math 106 – Exam #2 - Review Problems

Math 106 – Exam #2 - Review Problems. 1. (a) (b) (c) (d) (e) (f) (g). Fourteen different candy bars (Snicker’s, Milky Way, etc.) are available for a mother to give to her daughter.

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Math 106 – Exam #2 - Review Problems

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  1. Math 106 – Exam #2 - Review Problems 1. (a) (b) (c) (d) (e) (f) (g) Fourteen different candy bars (Snicker’s, Milky Way, etc.) are available for a mother to give to her daughter. How many ways can the mother give the daughter exactly four of the candy bars? C(14,4) = 1001 How many ways can the mother give the daughter exactly four or five of the candy bars? C(14,4) + C(14,5) = 1001 + 2002 = 3003 How many ways can the mother give the daughter more than three but less than eight of the candy bars? C(14,4) + C(14,5) + C(14,6) + C(14,7) = 1001 + 2002 + 3003 + 3432 = 9438 How many ways can the mother give the daughter more than ten of the candy bars? C(14,11) + C(14,12) + C(14,13) + C(14,14) = 364 + 91 + 14 + 1 = 470 How many ways can the mother give the daughter at least ten of the candy bars? C(14,10) + C(14,11) + C(14,12) + C(14,13) + C(14,14) = 1001 + 364 + 91 + 14 + 1 = 1471 How many ways can the mother give the daughter less than six of the candy bars, including the possibility of giving none? C(14,0) + C(14,1) + C(14,2) + C(14,3) + C(14,4) + C(14,5) = 1 + 14 + 91 + 364 + 1001 + 2002 = 3473 How many ways can the mother give the daughter at most six of the candy bars, including the possibility of giving none? C(14,0) + C(14,1) + C(14,2) + C(14,3) + C(14,4) + C(14,5) + C(14,6) = 6476

  2. (h) (i) (j) How many ways can the mother give the daughter some but not all of the candy bars? 214– 2 = 16,382 How many ways can the mother give the daughter more than three of the candy bars? 214– [C(14,0) + C(14,1) + C(14,2) + C(14,3)] = 214– 470 = 15,914 How many ways can the mother give the daughter at most 11 of the candy bars, including the possibility of giving none? 214– [C(14,12) + C(14,13) + C(14,14)] = 214– 106 = 16,278 Consider the equation x + y + z = 3 . 2. (a) (b) How many solutions in non-negative integers are there? 5! —— = 10 2! 3! Write each possible solution in non-negative integers together with either a sequence of slashes and “c”s or a sequence of “u”s and “b”s to represent that solution. x = 3, y = 0, z = 0 c c c / / x = 2, y = 0, z = 1 c c / / c x = 0, y = 2, z = 1 / c c / c x = 0, y = 3, z = 0 / c c c / x = 1, y = 2, z = 0 c / c c / x = 0, y = 1, z = 2 / c / c c x = 0, y = 0, z = 3 / /c c c x = 1, y = 0, z = 2 c / / c c x = 2, y = 1, z = 0 c c / c / x = 1, y = 1, z = 1 c / c / c

  3. 3. (a) (b) (c) (d) Fourteen identical candy bars (all Snicker’s) are available for second-grade teacher Rachael to distribute either to the nine other teachers in the lounge or to the 21 second-graders in her classroom. How many ways can Rachael distribute the candy bars among the other teachers in the lounge excluding herself? x1 + x2 + … + x8 + x9 = 14 non-negative integers 22! ——— = 319,770 8! 14! How many ways can Rachael distribute the candy bars among the other teachers in the lounge and herself? x1 + x2 + … + x9 + x10 = 14 non-negative integers 23! ——— = 817,190 9! 14! How many ways can Rachael distribute the candy bars among the second-graders in her classroom? x1 + x2 + … + x20 + x21 = 14 non-negative integers 34! ——— = 1,391,975,640 20! 14! How many ways can Rachael distribute the candy bars among the second-graders in her classroom and herself?

  4. x1 + x2 + … + x21 + x22 = 14 non-negative integers 35! ——— = 2,319,959,400 21! 14! (e) (f) (g) (h) How many ways can Rachael distribute the candy bars among the other teachers in the lounge, if she keeps exactly one for herself and guarantees that each teacher in the lounge gets at least one? x1 + x2 + … + x8 + x9 = 13 positive integers 12! ——— = 495 8! 4! x1 + x2 + … + x8 + x9 = 4 non-negative integers How many ways can Rachael distribute the candy bars among the second-graders in her classroom, if she keeps exactly one for herself and guarantees that each second-grader gets at least one? zero (0), since this is impossible to do How many ways can Rachael distribute the candy bars among the other teachers in the lounge excluding herself, if she identifies one teacher in the lounge who will get at least five? x1 + x2 + … + x8 + x9 = 14 5  x9 x1 + x2 + … + x8 + x9 = 9 non-negative integers 17! ——— = 24,310 8! 9! How many ways can Rachael distribute the candy bars among the second-graders in her classroom, if she identifies one second-grader in her classroom who will get at most six? x1 + x2 + … + x20 + x21 = 14 x21 6 x1 + x2 + … + x20 + x21 = 14 7  x21

  5. x1 + x2 + … + x20 + x21 = 7 non-negative integers 27! ——— = 888,030 20! 7! 1,391,975,640 – 888,030 = 1,391,087,610

  6. 3.-continued (i) (j) (k) (l) How many ways can Rachael distribute the candy bars among the other teachers in the lounge excluding herself, if she identifies one teacher in the lounge who will get at least three and identifies another teacher in the lounge who will get at least two? (ANSWERS ON NEXT SLIDE) How many ways can Rachael distribute the candy bars among the second-graders in her classroom, if she identifies one second-grader in her classroom who will get at least four and identifies another second-grader in her classroom who will get at most two? (ANSWERS ON NEXT SLIDE) How many ways can Rachael distribute the candy bars among the other teachers in the lounge excluding herself, if she identifies one teacher in the lounge who will get at least four but at most eight? (ANSWERS ON NEXT SLIDE) How many ways can Rachael distribute the candy bars among the second-graders in her classroom, if she identifies one second-grader in her classroom who will get at least four and identifies another second-grader in her classroom who will get at most eight? (ANSWERS ON NEXT SLIDE)

  7. 3.-continued (i) (j) (k) (l) x1 + x2 + … + x8 + x9 = 14 3  x8 & 2  x9 x1 + x2 + … + x8 + x9 = 9 non-negative integers 17! ——– = 24,310 8! 9! x1 + x2 + … + x20 + x21 = 14 4 x20 & x21  2 x1 + x2 + … + x20 + x21 = 14 4 x20 x1 + x2 + … + x20 + x21 = 14 4 x20 & 3  x21 30! ——— = 30,045,015 20! 10! 27! ——— = 888,030 20! 7! 30,045,015 – 888,030 = 29,156,985 x1 + x2 + … + x8 + x9 = 14 4  x9  8 x1 + x2 + … + x8 + x9 = 14 4  x9 x1 + x2 + … + x8 + x9 = 14 4  x9 & 9  x9 18! ——— = 43,758 8! 10! 13! ——— = 1287 8! 5! 43,758 – 1287 = 42,471 x1 + x2 + … + x20 + x21 = 14 4 x20 & x21  8 x1 + x2 + … + x20 + x21 = 14 4 x20 x1 + x2 + … + x20 + x21 = 14 4 x20 & 9  x21 30! ——— = 30,045,015 20! 10! 21! ——— = 21 20! 1! 30,045,015 – 21 = 30,044,094

  8. (m) How many ways can Rachael distribute the candy bars among the second-graders in her classroom, if she identifies one second-grader in her classroom who will get at least four but at most ten? x1 + x2 + … + x20 + x21 = 14 4 x21 x1 + x2 + … + x20 + x21 = 14 4 x21 & 11  x21 x1 + x2 + … + x20 + x21 = 14 4 x21  10 30! ——— = 30,045,015 20! 10! 23! ——— = 1771 20! 3! 30,045,015 – 1771 = 30,043,244 4. (a) (b) (c) A child likes two kinds of candy bars: Milky Way bars and Snicker’s bars. How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row? 220 = 1,048,576 How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that candy bars of the same type are never adjacent? 2 How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that Milky Way bars are never adjacent? F(20) = 17,711

  9. 4.-continued (d) (e) (f) (g) How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that Snicker’s bars are never adjacent? F(20) = 17,711 How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that at least two Milky Way bars are adjacent? 220– F(20) = 1,048,576 – 17,711 = 1,030,865 How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that at least two candy bars of the same type are adjacent? 220– 2 = 1,048,574 How many ways can 20 candy bars, each one either a Milky Way bar or a Snicker’s bar, be arranged in a row so that at least two Snicker’s bars are adjacent? 220– F(20) = 1,048,576 – 17,711 = 1,030,865

  10. (h) (i) (j) (k) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row? 20! ——— = 167,960 11! 9! How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that candy bars of the same type are never adjacent? 0 How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that Milky Way bars are never adjacent? 0 How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that Snicker’s bars are never adjacent? C(12,9) = 220

  11. 4.-continued (l) (m) (n) (o) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that the Milky Way bars are all adjacent? 10 How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that the Snicker’s bars are all adjacent? 12 How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that at least two candy bars of the same type are adjacent? 20! ——— – 0 = 167,960 11! 9! How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that at least two Snicker’s bars are adjacent? 20! ——— – C(12,9) = 167,960 – 220 = 167,740 11! 9!

  12. (p) (q) How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that at least two Milky Way bars are adjacent? 20! ——— – 0 = 167,960 11! 9! How many ways can 11 Milky Way bars and 9 Snicker’s bars be arranged in a row so that candy bars of the same type are all adjacent? 2

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