How can choosing players be equivalent to choosing non-players?

"And in the above example, we see that ( 4 0 ) is equal to ( 4 4 ) , and that ( 4 1 ) is equal to ( 4 3 ) . Why is this? Well, you can look back at the formula for ( n k ) and see how it works out algebraically. But it’s good to have an intuitive feel for it as well. Here’s how I think of it. Go back to the Davies kids and the Wii. We said there were three different ways to choose 2 kids to play on the Wii first after school. In other words, ( 3 2 ) = 3 . Very well. But if you think about it, there must then also be three different ways to leave out exactly one kid. If we change what we’re counting from “combinations of players” to “combinations of non-players” — both of which must be equal, since no matter what happens, we’ll be partitioning the Davies kids into players and non-players — then we see that ( 3 1 ) must also be 3. And this is true across the board. If there are ( 500 4 ) different lineups of four movies, then there are the same number of lineups of 496 movies, since ( 500 4 ) = ( 500 496 ) . Conceptually, in the first case we choose a group of four and show them, and in the second case we choose a group of four and show everything but them."