
Change combinatorix
So i know I need to use combinatorix for this, but I'm not exactly sure how.
4.3.19 In how many ways can you make change for a dollar, using pennies, nickels, dimes, quarters, and halfdollars? For example, 100 pennies is one way; 20 pennies, 2 nickels, and 7 dimes is another. Order doesn't matter.
Thanks in advance!

Did you try generating functions?
Let $\displaystyle a_n$ be the number of nonnegative solutions to,
$\displaystyle x_1+5x_2+10x_3+25x_4+50x_5=n$.
Create the generating function,
$\displaystyle f(x) = a_0+a_1x+a_2x^2+...$.
Now, realize that,
$\displaystyle g_1(x)g_2(x)g_3(x)g_4(x)g_5(x) = f(x)$
Where,
$\displaystyle g_1(x)=1+x+x^2+...$
$\displaystyle g_2(x)=1+x^5+x^{10}+...$
$\displaystyle g_3(x)=1+x^{10}+x^{20}+...$
$\displaystyle g_4(x)=1+x^{25}+x^{50}+...$
$\displaystyle g_5(x)=1+x^{50}+x^{100}+...$
Thus, by geometric series,
$\displaystyle f(x) = \frac{1}{1x}\cdot \frac{1}{1x^5}\cdot \frac{1}{1x^{10}}\cdot \frac{1}{1x^{25}}\cdot \frac{1}{1x^{50}}$
Does that work out?

There is no easy way to do this problem!
The most efficient way is by generating functions.
Can you expand the following expression:
$\displaystyle \left( {\sum\limits_{k = 0}^{100} {x^k } } \right)\left( {\sum\limits_{k = 0}^{20} {x^{5k} } } \right)\left( {\sum\limits_{k = 0}^{10} {x^{10k} } } \right)\left( {\sum\limits_{k = 0}^4 {x^{25k} } } \right)\left( {\sum\limits_{k = 0}^2 {x^{50k} } } \right)$?
If you can then the coefficient of $\displaystyle x^{100}$ is the answer to the question.

The highest term I got when multiplying all of that together was (3x^19)/(20(x^20x^15+x^10x^5+1)). Did I do something wrong?

I'm still stuck on this one. I don't get an x^100 term when I do the product of the geometric series. What do I do?