I know that when you multiply out $(1+x)^n$, the coefficient of $x^a$ tells you how many ways you can pick a of the brackets to use the x from to make $x^a$. I was wondering whether there is a meaning like this when you are multiplying brackets which do not have the same powers of x, i.e. $(1+x^a)(1+x^b)(1+x^c)...$
The reason I'm asking is because I read that the number of ways that you can make 10 pence out of 10p, 5p, 2p and 1p (which is 11 ways) is the coefficient of the expansion
$$ \frac{1}{(1-x^{10})(1-x^5)(1-x^2)(1-x)} $$
which I don't really get either...
Thanks in advance for any replies!
$$\frac{1}{1-x^{10}}=1+x^{10}+x^{20}+x^{30}+\cdots$$ This represents the number of ways to make $n$ cents out of 10p coins. There is 1 way to make $n=0$ (with zero coins), 0 ways to make $n=1,2,\ldots,9$, 1 way to make $n=10$ (with one coin), etc.
$$\frac{1}{1-x^{5}}=1+x^5+x^{10}+x^{15}+\cdots$$ This is a similar series to count how many ways there are to make $n$ cents out of 5p coins.
Now, multiply them together to get $$\frac{1}{(1-x^{10})(1-x^5)}=(1+x^{10}+x^{20}+x^{30}+\cdots)(1+x^5+x^{10}+x^{15}+\cdots)=$$ $$=1+x^5+2x^{10}+2x^{15}+3x^{20}+\cdots$$ Now, this counts how many ways to make $n$ cents out of 5p and 10p coins. $n=0$ has one way (no coins), $n=1,2,3,4$ has no ways, $n=5$ has one way, ... $n=10$ has two ways (one 10 coin or two 5p coins). The coefficient 2 comes from adding $1\cdot x^{10}$ (zero 10p and two 5p) and $x^{10}\cdot 1$ (one 10p and zero 5p).
The OP's expression is the same thing with 2 more coins.