This is a combination or groups question, but you don't need to use any complex factorials here if you don't want to.

Let's label our doors 1, 2, and 3.

For the first door, we can choose any of the 3.

Now, for the second door, we've got two remaining doors to choose from.

So we have 3*2=6.

BUT since we're talking about groups, the order doesn't matter. A group consisting of Doors 1 and 3 is the same as Doors 3 and 1--we don't want to double count them.

So simply divide it in half: \(6/2 = 3\)

If you prefer to use the combination formula: \(3C2 = \frac{3!}{{(3-2)!2!}}=\frac{3!}{{1!2!}}=\frac{6}{2}=3\)

Answer: B