Final answer:
To prove the combination formula c(n, r) = n! / ((n-r)! r!), we consider the total number of arrangements of n elements (n!), adjust for the irrelevant arrangements of n - r leftover elements by dividing by (n-r)!, and then correct for the overcounting of r element arrangements by dividing by r!.
Step-by-step explanation:
To prove the combination formula c(n, r) = n! / ((n-r)! r!), let's first understand what a combination is. A combination refers to the selection of r elements from a set of n elements without regard to the order of the elements. In terms of series expansions, we can link it to the Binomial theorem, but for this proof, we'll focus on the counting principle.
First, consider the number of ways to arrange n elements in a sequence, which is n!. If we are only interested in arranging r out of those n elements, we first select r elements—there are c(n, r) ways to do this.
Now, for each of these combinations, there are r! ways to order the r elements. However, since we do not care about the order in a combination, we have overcounted by a factor of r!. Therefore, we divide by r! to correct this. Lastly, we consider the remaining n - r elements that are not part of the combination. There are (n-r)! ways to arrange these, which we also need to account for by dividing, as these arrangements are not relevant to our combination of r elements.
Putting it all together, the number of combinations is obtained by multiplying the number of ways to pick r out of n elements (n!) by the number of irrelevant arrangements of the remaining elements ((n-r)!), and then dividing by the overcounting from the ordering of the r elements themselves (r!), which gives us the formula c(n, r) = n! / ((n-r)! r!).