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Guides d'étude > Precalculus II

Solutions for Counting Principles

Solutions to Try Its

1. 7 2. There are 60 possible breakfast specials. 3. 120 4. 60 5. 12 6. P(7,7)=5,040P\left(7,7\right)=5,040 7. P(7,5)=2,520P\left(7,5\right)=2,520 8. C(10,3)=120C\left(10,3\right)=120 9. 64 sundaes 10. 840

Solutions of Odd-Numbered Exercises

1. There are m+nm+n ways for either event AA or event BB to occur. 3. The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word "or" usually implies an addition problem. The word "and" usually implies a multiplication problem. 5. A combination; C(n,r)=n!(n−r)!r!C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!} 7. 4+2=64+2=6 9. 5+4+7=165+4+7=16 11. 2×6=122\times 6=12 13. 103=1000{10}^{3}=1000 15. P(5,2)=20P\left(5,2\right)=20 17. P(3,3)=6P\left(3,3\right)=6 19. P(11,5)=55,440P\left(11,5\right)=55,440 21. C(12,4)=495C\left(12,4\right)=495 23. C(7,6)=7C\left(7,6\right)=7 25. 210=1024{2}^{10}=1024 27. 212=4096{2}^{12}=4096 29. 29=512{2}^{9}=512 31. 8!3!=6720\frac{8!}{3!}=6720 33. 12!3!2!3!4!\frac{12!}{3!2!3!4!} 35. 9 37. Yes, for the trivial cases r=0r=0 and r=1r=1. If r=0r=0, then C(n,r)=P(n,r)=1.C\left(n,r\right)=P\left(n,r\right)=1\text{.\hspace{0.17em}} If r=1r=1, then r=1r=1, C(n,r)=P(n,r)=nC\left(n,r\right)=P\left(n,r\right)=n. 39. 6!2!×4!=8640\frac{6!}{2!}\times 4!=8640 41. 6−3+8−3=86 - 3+8 - 3=8 43. 4×2×5=404\times 2\times 5=40 45. 4×12×3=1444\times 12\times 3=144 47. P(15,9)=1,816,214,400P\left(15,9\right)=1,816,214,400 49. C(10,3)×C(6,5)×C(5,2)=7,200C\left(10,3\right)\times C\left(6,5\right)\times C\left(5,2\right)=7,200 51. 211=2048{2}^{11}=2048 53. 20!6!6!8!=116,396,280\frac{20!}{6!6!8!}=116,396,280

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  • Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. License: CC BY: Attribution.