homework 9 solution - 16 11 12 10626 17 2520 18 20 4(3 3(2...

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Selected Hints/Answers to HW 9 October 30, 2012 Note: 1. There is a possibility of typos; in case you notice one, let me know. 2. These are just the answers; you must show work in your homework/quiz to get full credit. Section 6.2 29. R (2 , n ) is the minimum number of people gathered together such that there are at least 2 mutual friends or n mutual enemies. Clearly R (2 , n ) n but why is R (2 , n ) = n ? 30. Is it possible to switch the notion of friend/enemy? 36. Consider the number of computers that each computer can directly be connected to. What are some possible values for such numbers? 40. This example was discussed in class. 42. a) Yes. b) Yes. c) Yes (but one has to be careful as just Pigeonhole Principle isn’t enough. 44. Similar to example discussed in class. 47. Use hint for part b). Section 6.5 1. 243. 4. 6 7 . 5. 5 3 . 6. 21. 8. ( 21+12 - 1 12 ) = ( 32 12 ) . 1
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9. a) 1716. b) 50388. c) 2629575. d) 330. e) 9724. 10. a) ( 17 12 ) . b) ( 41 36 ) . c) ( 17 12 ) . d) ( 29 24 ) - ( 26 21 ) . e) ( 21 16 ) . f) ( 20
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Unformatted text preview: ( 16 11 ) . 12. 10626. 17. 2520. 18. 20! 4!(3!) 3 (2!) 3 . 20. 364. 21. ( 14 6 ) . 23. 12! (2!) 6 . 24. If you assume that the first box has 1 object, the second one 2, etc., then the number of ways to distribute the objects is 15! 1!2!3!4!5! . However, the way the question is phrased, it only seems to say one of the boxes (not necessarily the first) has 1 object, some other box has 2 objects, etc. and in this case the number of ways to distribute the objects is 5!15! 1!2!3!4!5! . 25. ( 24 19 )-6 ( 14 9 ) . 26. 6 ( 8 4 ) . 32. 6! 2! . 33. 63. 34. 7! 3!3! + 6! 3!3! + 2 · 6! 3!2! + 2 · 5! 3! + ! 2!2! + 2 · 5! 3!2! = 370. 38. a) 40! (10!) 4 . b) 40! 4!(10!) 4 . 39. 12! 4!3!5! . 42. 52! 13! 4 . 48. Observe that we do not consider the ordering of the elements inside the boxes, and hence they can be treated as indistinguishable. 2...
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