# A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

There are 6 people. A game that requires one pair vs another pair to play is to take place - how many unique games can take place?

My Way: 6C2 pairs = 15, and 15C2 distinct matche between each possible pair = 105 ways.

Others: 6C4 distinct participants, and for each 3 ways to hold a match = 15 * 3 = 45.

Why's my way going wrong? I can list out 15 unique pairs and have them play each other in 105 ways?

Your pairs overlap. It's no good taking $$(p_1, p_2)$$ and $$(p_1,p_3)$$ as your two pairs, right?

To repair your calculation: Having chosen one pair, there are now $$4$$ people left to choose from, and $$\binom 42=6$$. So you get $$\frac {15\times 6}2=45$$ as you should.

With the tennis season heating up, we can look at it like arranging doubles tennis matches.

$$4$$ individuals can be selected in $$\binom64 = 15$$ ways

and the tallest among them can be paired with any of the remaining $$3$$,

thus $$15\times3 = 45$$ ways

A more pedestrian approach: you need to select 4 people out of 6, for the first one you have six possibilities, second: five etc. But then you need to rule out some permutations: one within each pair, and one between pairs. So in total: $$\frac{6\cdot5\cdot4\cdot3}{2\cdot2\cdot2} = 45$$