Probability of forming a 3-senator committee If the Senate has 47 Republicans and 53 Democrats,
in how many ways can you form a 3-senator committee 
in which neither party holds all 3 seats?

The solution says that:
You can choose one Democrat, one Republican, and one more senator from either party. We can make these choices, in that order, in $53\cdot 47\cdot 98$ ways.
But then we've counted each possible committee twice, since any given committee can be arranged in the order Democrat-Republican-Third Person in two different ways (depending on which member of the majority party on the committee is chosen as the Third Person).

How are there two different ways to arrange the Democrat-Republican-Third committee based on the third person chosen? I only see one possible way.

Hint: find 
[Number of ways of choosing arbitrary set of senators] - 
[number of ways of choosing 3 Democrats] - 
[number of ways of choosing 3 Republicans].  
Finding each of these 3 quantities is a standard probability task (think combininations, permutations, etc.).
NOTE: I'm ignoring the given solution, since I think it makes the problem look more difficult than it actually is.
