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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?

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3 Answers 3

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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.

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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

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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: $$ \frac{6\cdot5\cdot4\cdot3}{2\cdot2\cdot2} = 45 $$

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