The official pokémon video game championship (VGC) tournaments allow you to have a team of 6 pokémon, and at the beginning of the battle you see a team preview of your opponent's 6 pokémon to decide which 4 out of your 6 you will battle with.
The VGC ruleset uses the doubles format, meaning each player will send out 2 pokémon to battle with at the same time, until all 4 of a trainer's pokémon faints (or until time runs out.) What this means for the math problem is that the order sort of matters.
To make that last part clearer, let's say I have a team of pokémon A through F:
First, I decide I will be using A, B, C, and F after seeing the team preview. However, I still have to decide which 2 I will start with, which I can designate with parentheses in this example.
(AB)CF is not different from (BA)FC since I would be starting with the same two pokémon either way. However, (AF)BC could change the result of the battle if I start with the wrong pokémon.
How many different ways could I choose my pokémon for a battle following the VGC format?
First choose the four Pokémon to use in the battle, then choose the two Pokémon to start with. The number of ways in which to do this equals:
$${6 \choose 4}{4 \choose 2} = \frac{6!}{4!2!} \frac{4!}{2!2!} = 15 \cdot 6 = 90$$
Alternatively, you can also look at it this way. First order the six Pokémon, which you can do in $6! = 720$ ways. The first two are the ones you start with, the second two are the ones you will use later in the battle. The order in which the first, the second and the third two Pokémon are selected does not matter, so we must divide by $2$ for each of these couples. We thus get:
$$\frac{6!}{2 \cdot 2 \cdot 2} = \frac{720}{8} = 90$$