What does the binomial theorem applied to integers count? I have a homework problem that I'm stuck on, but I feel like if I could get help on this one component (not the whole problem) I could make progress. Also, I'll note preemptively that I tried searching for this, but I really couldn't find anything so if this is a duplicate I apologize in advance.
The problem asks: Evaluate $\sum_{k=0}^n {n \choose k} a^{n-k} (-b)^k$ for $a \geq b > 0$ using an involution. To save some searching, an involution is a function $I$ on $X$ such that $I \circ I = id_X$.
My actual question is, what does $\sum_{k=0}^n {n \choose k} a^{n-k} b^k$ count? That is, what does the binomial theorem count when evaluated for positive integers $a,b$? My intuition is that if I understood this I might be able to define an involution that I can use.
It's worth noting that the answer is easy if you just apply the binomial theorem with $x = a, y = -b$, but using an involution is the key component of the problem.
 A: $$
(a+b)^4 = a^4+4a^3 b + 6a^2b^2 + 4ab^3 + b^4
$$
The coefficients $1,4,6,4,1$ count combinations, and in order to see them, I will make the four $a$s and four $b$s in $(a+b)^4$ distinguishable from each other, writing $(a+b)^4$ as $(a_1+b_1)(a_2+b_2)(a_3+b_3)(a_4+b_4)$.
\begin{align}
{} \\
& (a_1+b_1)(a_2+b_2)(a_3+b_3)(a_4+b_4) \\[10pt]
= {} & \underbrace{a_1a_2a_3a_4}_{\text{1 term, equal to $a^4$}} + \underbrace{a_1a_2a_3b_4 + a_1a_2b_3a_4 + a_1b_2a_3a_4+b_1a_2a_3a_4}_{\text{4 terms, each equal to $a^3b$}} \\[6pt]
{} + {} & \overbrace{a_1a_2b_3b_4 + a_1b_2a_3b_4+b_1a_2a_3b_4 + a_1b_2b_3a_4+b_1a_2b_3a_4 + b_1b_2a_3a_4}^{\text{6 terms, each equal to $a^2b^2$}} + \cdots\cdots \\  {}
\end{align}
So the $\dbinom 4 2 = 6$ term counts the six ways to choose two indices from among $1,2,3,4$.  The two chosen indices are $b$s and the others are $a$s, or you can view it the other way, the chosen ones being $a$s and the others $b$s.  And similarly for $\dbinom n k$ for other values of $n$ and $k$.
