Why is the combination formula (without replacement) not just $\frac{n(n - 1)(n - 2) \cdots (n - r + 1)}{r!}$? 
Why is the combination formula (without replacement) not just $\frac{n(n - 1)(n - 2) \cdots (n - r + 1)}{r!}$?

I tried many examples. Both the title formula and the standard combination formula gave the same answers. At first, I thought it was because of something about $\binom{n}{r}$ and $\binom{n}{n - r}$, but those examples didn't seem to matter.
Ex: from $\binom{5}{2}$, from $\binom{5}{3}$ gave the same answers from both formulas and  from $\binom{20}{17}$, from $\binom{20}{3}$ gave the same answers from both formulas)
 A: Great question! The answer is that mathematicians know and love both formulas, and will use whichever one is convenient for the purpose at hand. (It is amazingly useful to have alternative expressions for the same quantity; that's why we spend so much time learning identities.)
The formula $$C(n,r)=\frac{n!}{r!(n-r)!}$$ has the advantage of being concise and easy to remember, and of making it plain to see that $C(n,r)=C(n,n-r)$; it generalizes to the multinomial formula $$C(n;r_1,r_2,\dots,r_k)=\frac{n!}{r_1!r_2!\cdots r_k!};\ r_1+r_2+\cdots+r_k=n.$$
On the whole, your formula $$C(n,r)=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$$
is the better of the two. First, it's easier to compute, though for computation the form $$C(n,r)=\frac n1\times\frac{n-1}2\times\frac{n-2}3\times\cdots\times\frac{n-r+1}r$$ may be even better. More importantly, your formula makes it apparent that $C(n,r)$ makes sense even when $n$ is not a natural number; for example, $$C(n,3)=\frac{n(n-1)(n-2)}6=\frac16n^3-\frac12n^2+\frac13n$$ is just a cubic polynomial, defined for all real (or complex) values of $n$. Of course such generalized "combination numbers" as $C(-\frac12,r)$ are not useful for counting poker hands, but they are just what you need if you want to use the Binomial Theorem to expand a binomial with a negative or fractional exponent, such as $$(1+x)^{-\frac12}=\sum_{r=0}^\infty C(-\frac12,r)x^r=1-\frac12x+\frac38x^2-\frac5{16}x^3+\cdots.$$
A: I'll agree that if one is learning elementary stuff then for most of the times instead of $\frac{n!}{r!(n-r)!}$, the equivalent formula $\displaystyle\frac{n⋅(n-1)⋅(n-2)\cdots(n-r+1)}{r!}$ comes much more handy.
Example:

*

*$^8C_5=\binom85=\binom83=\displaystyle\frac{8⋅7⋅6}{1⋅2⋅3}=56\quad$ which is easier than: $\quad\displaystyle\frac{8!}{5!3!}=\frac{40320}{(120)(6)}$
Obviously that's why one is also taught the properties (such as one you already observed) to make such tasks easier like: $\binom{n}{r}=\binom{n}{n-r}; \ \binom nr+\binom n{r-1}=\binom{n+1}r; \ $ etc.

But once we move onto stuff like calculating $\displaystyle\binom{43}{26}$ now I would rather prefer the former over latter because, first, it's neat, second, latter isn't really helpful. 
Other cases may be like: $\displaystyle\frac{26!}{7!8!8!}$, etc.

Surprisingly though I often see that some students don't see that the latter may be derived from former and thus would instead eveytime expand the numerator to be surprised to see something "big" getting cancelled from the denominator. Funny but sad too and add to that some confident teachers who are too sure of their students' ability to observe these things. Unlike you though, many don't. They'll basically always solve in this way: $$\displaystyle\binom85=\binom83=\frac {8!}{5!3!} \ (\text{LOL})= \frac{8.7.6.5!}{5!3!}=\frac{8.7.6}{6}=56 $$
