There are 6 people. I have to pick 3 teams with two members..how many selections are possible Is it $\dbinom{6}{2}*\dbinom{4}{2}*\dbinom{2}{2}$ or just $\dbinom{6}{2}$ or something else?
There are $\binom{6}{2}$ ways to pick the first team, $\binom{4}{2}$ ways to pick the second team, and $\binom{2}{2}$ ways to pick the third team.  However, we have over counted since the selection $\{a, b\}, \{c, d\}, \{e, f\}$ produces the same teams as the selection $\{c, d\}, \{e, f\}, \{a, b\}$.  Since there are three teams, we must divide the $\binom{6}{2}\binom{4}{2}\binom{2}{2}$ ways we can select the teams by the $3!$ ways we can select identical teams, giving 
$$\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$$ 
distinct ways of selecting three teams each consisting of two people selected from a group of six people.
Primary assumption: 3 teams, each with 2 members (not entirely clear in the original question).

Labelling the members {a, b, c, d, e, f}

Assuming that two people could be in the same team in more than one combination:

(6C2) * (4C2) * (2C2) / 3! = 15

would be correct, allowing for these to be valid:


{a, b} {c, d} {e, f} 
{a, b} {c, e} {d, f} 
{a, b} {c, f} {d, e} 


However, if each person can only be teamed with another person once, 
then it only matters how many groups we can make with the first person {a}:

{a, b}, {a, c}, {a, d}, {a, e], {a, f}

(6-1) = 5

to get the number of possible groupings:


{a, b} {c, d} {e, f} 
{a, c} {b, e} {d, f} 
{a, d} {b, f} {c, e} 
{a, e} {b, d} {c, f}
{a, f} {b, c} {d, e}