```1. 5.4.10
C(32,4); C(27,4); C(17,4)

2. 5.4.22
C(17,2)+C(14,2)+C(11,2)+C(8,2)+C(5,2)+1

3. 5.4.48 solved at end of chapter

4. 5.4.54
choose empty boxes: C(n,m). Now distribute balls among other boxes. Final answer:
C(n,m)*C(r+n-m-1,r)

5. 5.5.14,d,f For d: 3n*4**(n-1), which I get by differentiating (1+3x)**n. The
derivative of its power series expansion, evaluated at x = 1, is the series being
sought. f: integrate (1+x)**n from 0 to 1. Termwise you get the desired
expression. Integrating the expression directly you get: ((1/n+1)2**(n+1))-(n+1)

6. 5.5.22 Differentiate (1+x)**n termwise, then set x=-1, which puts odd, even
terms on opposite sides.

7. 5.5.26  Take n red things and m blue things, and choose any subset of size m.
Each term in the sum describes a pattern of ways that this can be done with 0 reds
+ m blues, 1 red m-1 blues, etc. So the sum is: C(m+n,m)

8. 5.5.34 A term with a bunch of xis raised to various powers describes the ways
in which xis can be chosen from the various factors, when (x1+..xn)**n is written
out as a massive product of factors.

9. 6.1.2  (a) (x1+x1**2+x1**3+x1**4)**5; others are similar

10. 6.1.6 (x1+x1**2+x1**3+x1**4 +..)**5 seek coefficient of x**r

11. 6.1.22  Imagine solution to equation in unary notation. Every "1" in e1 must
have a corresponding 1 in the other terms as well. Thus: the last term represents
e1, since each value for e1 must be "spent" for e2,e3,e4 as well. Next to last
term represents e2, and so forth.
```

-- RobbieMoll - 30 Nov 2006

Topic revision: r1 - 2006-11-30 - RobbieMoll    Copyright © 2008-2019 by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
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