There are 6 people. I have to pick 3 teams with two members..how many selections are possible

Question

Is it $\dbinom{6}{2}*\dbinom{4}{2}*\dbinom{2}{2}$ or just $\dbinom{6}{2}$ or something else?

Answer 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.

Answer 2

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}