In the book "Discrete Mathematics and Its Applications", it defined the combinatorial proof to me. And then it illustrated how to use it to proof C(n,r)=C(n,n-r).
The segment of its text is:
...Suppose that S is a set with n elements. The function that maps a subset A of S to $\overline{A}$ is a bijection between subsets of S with r elements and subsets with n − r elements....
I can't get any idea about how to map two sets of two different cardinalities with a bijection. Such as, n=10 and r=4, it meants that we should establish a bijection from a set whose cardinality is 4 to another set whose cardinality is 10-4=6? I think it is impossible. So where have I misunderstanded?
$$\begin{array}{rcl} \{1,2,3,4\}&\iff&\{5,6,7,8,9,10\}\\\{1,2,3,5\}&\iff&\{4,6,7,8,9,10\}\\\{1,2,3,6\}&\iff&\{4,5,7,8,9,10\}\\&\vdots\\ \{2,5,6,8\}&\iff&\{1,3,4,7,9,10\}\\&\vdots\end{array}$$
In general, a set $A$ corresponds to the set $A^c$. With the original set $A$ being four elements, the corresponding set $A^c$ is a six-element set here.
In the left column we have the list of all possible four-element subsets of $\{1,2,\dots,10\}$. In the right column we have the list of all possible six-element subsets of $\{1,2,\dots,10\}$. You should be able to see that the lists are exactly the same length. This is not to say that the element $\{1,2,3,4\}$ is in bijection with the element $\{5,6,7,8,9,10\}$... but rather that the set of sets of size four is in bijection with the set of sets of size six.
Explained a different way, "To choose four elements to have in your set out of ten options, you can instead choose which six elements to not have in your set."
In general, the principle of counting in two ways or "combinatorial proofs" you come up with a scenario that you can successfully count by different methods, each method resulting in a different expression. Assuming you counted correctly in each of those ways, then despite the resulting expressions appearing different since they both correctly counted the same scenario the expressions must be equal.
Here, since we can count how many 4-element subsets there are from a set with ten elements by picking the four elements in the subset and letting all other elements not be in the subset... giving the total count as $\binom{10}{4}$, and that we can also count how many 4-element subsets there are from a set with 10 elements by picking the six elements not in the subset and letting all other elements be in the subset... giving the count as $\binom{10}{6}$, we know that it must be the case that $\binom{10}{4}=\binom{10}{6}$, and by generalizing that $\binom{n}{r}=\binom{n}{n-r}$. This, without ever having to perform algebraic manipulations to show they are equal (even if it were trivial in this case).