5.4 #9,11,18,22,25,26a,49 <br>
5.5 #3,5,10,12,17 
<p>

5.4 #18. Select 14 balls from 100 red, 100 blue, 100 green.
We have to independently choose
<menu>
<li> how many red, from 0 to 100; 
<li> how many blue, from 3 to 10;
<li> how many green, from 0 to 100;
</menu>
such that it adds up to 14.
<p>
(1 + x + x^2 + ... + x^100) (x^3 + x^3 + ... + x^10) (1 + x + x^2 + ... + x^100)
<p>
Multiply this out, and take the coefficient of x^14.
<p>

#22. (1 + x + x^2 + x^3 + ...)^n means we make n choices of natural numbers,
and add them together. If we get 6 when we're done, then we added up n numbers
and got 6, which stars & bars tells us can be done as (n+5 choose 5).
<p>

#26a.
Let u = (x+x^2+x^3+x^4+x^5+x^6). Then 1/(1-u) = 1 + u + u^2 + u^3 + ...
<p>
The interpretation of u^k is that we're picking k numbers, in order, from 1..6,
and adding them together. So the coefficient of x^n in u^k is the number of
ways of breaking n into k numbers from 1..6.
<p>
Inside the sum 1 + u + u^2 + ..., we're doing it for all k. So the coefficient
of x^n is the number of ways of breaking n up into 1..6.

<p>
5.5 #10.
Those not divisible by 5 or 7 <br>
= all - (those divisible by 5) - (those divisible by 7) 
+ (those divisible by both) <br>
= 100 - 100/5 - floor(100/7) + floor(100/35) <br>
= 100 - 20 - 14 + 3 = 69.

<p>
#12. The squares or cubes = the squares + the cubes - both =
floor(1000 ^ 1/2) + floor(1000 ^ 1/3) - floor(1000 ^ 1/6) = 31 + 10 - 3 = 38.
("Both" is the sixth powers.)