How to simplify a sum of combinatorial values into a single value?

Question

Edit: I've tried to make this question more clear.

I know how algebra and calculus work, and can solve problems like these, where the point of the problem is to just simplify a term into its simplest form:

$$\dfrac{x^4 + x^4}{x^5} = \dfrac{2x^4}{x^5} = \dfrac{2}{x}$$

$$\int_0^y 2x = x^2 \bigr\rvert_0^y = y^2 - 0^2 = y^2$$

As in, I know what the middle steps are to get from the left-most side of the equation to the right-most side of the equation. But, I don't know what the middle steps of this equation are:

$$\sum_{r=0}^n \dfrac{n!}{r!(n-r)!} =\mathbf?=2^n$$

If this was a question on a test and I had to show my work, what would the middle steps be for this equation? Is there a way to solve it like you could solve the algebra and calculus problems above? I really hope this isn't a dumb question, but I just want to know and haven't been able to figure it out myself.

Answer 1

Well, the algebraic way I know to prove the statement you have given is by the binomial theorem; just expand out $(1+1)^n$ (Binomial theorem in spirit is still combinatorial though).

Intuitively though, one of them sums up the number of ways you can choose $r$ ingredients out of a list of $n$ things, and you sum over all $r$; that just gives you the number of subsets you could have, since you essentially include all possible subsets this way.

Answer 2

Think of this as a set, such as $\{\text{pork, cabbage, shrimp}\}$. For each ingredient, you can choose whether to include or not to include it. Therefore for $x$ ingredients, you have $2^x$ ways of making dumplings with those ingredients.