Binomial Expansion Is this is as heavy into combinatorics as we're going to go? Right, you wish. First off recall the binomial expansion formula: $$(a+b)^n = a^n + n a^{n-1}b + {n(n-1)\over 2} a^{n-2}b^2 + ... + b^n$$ (1.5) But what are all those terms in the middle? We write them in terms of combination notation $$\left(\!\!\! \begin{array}{c} n \\ m \end{array} \!\!\!\right) = \frac{n!}{m!(n-m)!}$$ (1.6) This means the "combination of n things taken m at a time". This is also sometimes written as ${^n}C_m$. Then we can write the binomial expansion as $$(a+b)^n = \sum_{m=0}^n \left(\!\!\! \begin{array}{c} n \\ m \end{array} \!\!\!\right) a^m b^{n-m}$$ (1.7) Also the notation "!" means "factorial. So for example $5! = 5\times4\times3\times2\times1$. This can be confusing when you want to write "how about that 5!", but mathematicians seldom get excited in papers. Well let's review how you derive this binomial expansion and what the right hand side means. Start off expanding $(a+b)(a+b) = aa + ab + ba + bb$. Let's now try $(a+b)(a+b)(a+b) = aaa+ (aab + aba + baa) + (bba+bab+abb) + bbb$. What's happening here is that this expansion generates all possible combinations a's and b's which I grouped together as above, so that you can easily write this as $a^3+3a^2b + 3 ab^2 + b^3$. The coefficients of this are simple cases of that general combination written above. In this case $(aab + aba + baa)$ represents then is the number of ways of choosing two $a$'s from a three possibilities. Another way of representing this is in terms of Pascal's triangle, but the factorial formula is usually easier to work with mathematically. I'm not going to go over a proof of this formula, but there are a lot of discussions of it. You've likely seen it before in any case. It's a variation on the idea of a permutation. For example if we have three things a, b, and c, then there are 6 permutations: abc, bac, cab, acb, bca, and cba. It is pretty easy to show that in general for $n$ things, there are $n!$ permutations of them. That's sort of the reason why $!$ appears three times in the combination formulation.