## 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: can’t 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 a’s, 3 b’s, and 3 c’s. 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 a’s, b’s,
c’s. (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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