I was thinking of a double counting proof of the binomial theorem for positive integers $x,y$. I want to verify if the following argument is correct:
Argument: Let us consider a group of $n$ boys. We want to form a committee of $k$ boys out of them and each person will be assigned a rank depending on whether the person is in the committee or not.
If the person is in the committee, he will be assigned one of $x$ positive ranks (for simplicity, say one of $1,2,\ldots,x$) and if he's not in the committee, he will be assigned one of $y$ negative ranks (for simplicity, say one of $-1,-2,\ldots,-y)$
Now, one way to do this is to first select a group of $k$ boys out of $n$ boys in $\binom nk$ ways and then assign the $k$ people the positive ranks in $x^k$ ways and assign the remaining people negative ranks in $y^{n-k}$ ways. By the rule of product, there are a total of $\sum_{k=0}^n\binom nk x^k y^{n-k}$ ways to do this.
Another way to count the same thing is to note that for every boy will be assigned one of $(x+y)$ ranks, so there are $(x+y)$ choices for each boy and by the rule of product, there are a total of $(x+y)^n$ ways.
Since we counted the same thing in two different ways, we conclude that both the expressions are equivalent.
Is my argument correct? Feedbacks and improvements are welcomed. Thanks!