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

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

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