This result is a common point of confusion, but it can be cleared up easily by considering both the algebra behind it and by looking at it conceptually.
First, let's start with the formula. We're basically trying to prove that the following equation holds true:
`((n),(k)) = ((n),(n-k))`
To do this, let's look at the formula for a combination:
`((n),(k)) = (n!)/(k!(n-k)!)`
So, let's show that the same holds true for n choose n-k:
`((n),(n-k)) = (n!)/((n-k)!(n-(n-k))!)`
Simplifying the denominator, we see that we get the same result as n choose k:
`= (n!)/((n-k)!k!) = ((n),(k))`
Therefore, algebraically, choosing 8 people out of 12 gives the same number of possibilities as choosing 4 people out of 12:
`((12),(8)) = ((12),(4))`
Now, let's talk about it more intuitively. Suppose that we're trying to make a group of 8 people chosen from 12. This situation is exactly the same as choosing how to exclude 4 people from a group of 12. It's like telling 8 people they can be on a committee is the same as telling 4 people that they can't.
I hope that helps!