Solutions to Try Its

1. 7

2. There are 60 possible breakfast specials.

3. 120

4. 60

5. 12

6. $P\left(7,7\right)=5,040$

7. $P\left(7,5\right)=2,520$

8. $C\left(10,3\right)=120$

9. 64 sundaes

10. 840

Solutions of Odd-Numbered Exercises

1. There are $m+n$ ways for either event $A$ or event $B$ 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; $C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!}$

7. $4+2=6$

9. $5+4+7=16$

11. $2\times 6=12$

13. ${10}^{3}=1000$

15. $P\left(5,2\right)=20$

17. $P\left(3,3\right)=6$

19. $P\left(11,5\right)=55,440$

21. $C\left(12,4\right)=495$

23. $C\left(7,6\right)=7$

25. ${2}^{10}=1024$

27. ${2}^{12}=4096$

29. ${2}^{9}=512$

31. $\frac{8!}{3!}=6720$

33. $\frac{12!}{3!2!3!4!}$

35. 9

37. Yes, for the trivial cases $r=0$ and $r=1$. If $r=0$, then $C\left(n,r\right)=P\left(n,r\right)=1\text{.}\hspace{0.17em}$ If $r=1$, then $r=1$, $C\left(n,r\right)=P\left(n,r\right)=n$.

39. $\frac{6!}{2!}\times 4!=8640$

41. $6 - 3+8 - 3=8$

43. $4\times 2\times 5=40$

45. $4\times 12\times 3=144$

47. $P\left(15,9\right)=1,816,214,400$

49. $C\left(10,3\right)\times C\left(6,5\right)\times C\left(5,2\right)=7,200$

51. ${2}^{11}=2048$

53. $\frac{20!}{6!6!8!}=116,396,280$