The order doesn't matter. It's the combination then.
[tex]k [/tex] objects can be chosen out of [tex] n [/tex] objects, when the order doesn't matter, in [tex] C(n,k)=\dfrac{n!}{k!(n-k)!} [/tex] ways.
So, the answer is [tex] C(6,2)=\dfrac{6!}{2!4!}=\dfrac{5\cdot6}{2}=15 [/tex] ways.