Urn without Replacement, all balls the same There's an urn with 12 balls: 5 white, 7 black. 4 balls are picked without replacement. 
What are the odds that all balls picked are white?
I know part of the answer involves the term $\frac{7}{12}*\frac{6}{11}*\frac{5}{10}*\frac{4}{9}$, but I also know that since order doesn't matter in the final result, then we must account for this. 
There's $\binom{5}{4}$ ways to choose 4 balls in a set of 5. 
So would my answer simply be this?
$\frac{\frac{7}{12}*\frac{6}{11}*\frac{5}{10}*\frac{4}{9}*\binom{5}{4}}{\binom{12}{4}}$
The numerator is that because the probability is one instance of picking 4 black balls, and the binomial coefficient accounts for all ways to do that. 
Then the denominator is that because it's dividing by all possible ways to pick 4 balls from 12 balls
 A: We may temporarily imagine that every ball has a unique number on it.
Now... we have the option to treat our sample space as the ways in which we draw four balls in sequence and we record the numbers of the balls that we drew, in which case we have $12\times 11\times 10\times 9$ possible equally likely results...
Or... we have the option to treat our sample space as the ways in which we draw four balls simultaneously where order doesn't matter and we record the set of the numbers we drew, in which case we have $\binom{12}{4}$ possible equally likely results...
It is a choice.  We have the option of picking whichever sample space we like, so long as it is a convenient choice to make.  We just need to remember which choice we made and run our calculations for the number of favorable cases with that in mind, also keeping in mind that in order to use the formula that the probability is the number of favorable cases divided by the total number of cases we must have picked a sample space which is equiprobable.

Having treated order as relevant, the probability (don't confuse the word probability with odds, they mean two different things) that all balls are white will be $\dfrac{5\times 4\times 3\times 2}{12\times 11\times 10\times 9}$ seen by direct application of rule of product, there are five choices for the first white ball, four choices for the second white ball, and so on...
Having treated order as irrelevant, the probability that all balls are white will be $\dfrac{\binom{5}{4}}{\binom{12}{4}}$.
You will notice that these both result in the same value once simplified.
