Homework Seven Quick Answers


1. 5.1.6  [n * (n-1)]


2. 5.1.10  23, that is (1+3)*(1+5) 1


3. 5.1.12 (5**2)*(26**2); (26**2)*(21**2)


4. 5.1.18 2*(26**3)(10**3); 
   (26 + 26**2 + 26**3)*(2*10 + 3*10**2 + 4*10**3)

5. 5.1.20   [(11*11) + (10*10)]/(21*21)


6. 5.1.24   10**8

7. 5.1.30 
   C(5,1)*9**4 + C(5,2)*9**3 + C(5,3)*9**2 + C(5,4)*9 + C(5,5)

8. 5.1.34   
   (n+1)**4 1

9. 5.1.42   
   I assume non-adjacent means: cant take each other. So I get:
   64*((49*55)+(28*58)+(4*60))

10. 5.2.2   
   P(26,10)

11. 5.2.6
   C(26,4)

12. 5.2.20
   (13)**4 x (12)**4 x  x 2**4 x 1
   
13. 5.2.24
   P(20,10)

14. 5.2.28
   C(7,4)

15. 5.2.70
   C(11,7)

16. 5.2.74
   2**(n-1) [will discuss in class..]

17. 5.3.2
   11!/(4!)(4!)(2!)

18. 5.3.30
`   Arrange the consonants first, in 4! ways.  The, for each of the 3 patterns of 
(separated) vowels (e.g. ei x i), these can be placed in 
5x4 + 4x3 + 3x2 + 2x1 ways. So answer is: 3x4!x[5x4 + 4x3 + 3x2 + 2x1]

19. 5.3.32
   Answer in text, on page 236. 

20. Show that (2**n x 3**n) divides (3n)!

   Take 3n objects consisting of 3 identical copies of each of n distinct 
objects.

21. Show that (n!)**(n+1) divides (n**2)! Look at the case of n = 3, and imagine 
that you are arranging 3 as, 3 bs, and 3 cs. All possible arrangements yields 
(n**2)!. Identical letters gives (n!)**n in denominator, and one more factor of n! 
in denominator is possible if we allow permutations among the groups of as, bs, 
cs. (Thus, aaabbcccb is identified with bbbaaccca)

22. Will do in class..

-- RobbieMoll - 30 Nov 2006

Topic revision: r1 - 2006-11-30 - RobbieMoll
 
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