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?

Solution 1:

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.