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I've come across the following problem in my textbook

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I tried solving this by figuring out how many choices there are for each position within a round and came up with the following.

$${8 \choose 2} \cdot {6 \choose 2} \cdot {4 \choose 2} \cdot {2 \choose 2}$$

My idea was that that are ${8 \choose 2}$ ways to pick the first pair, ${6 \choose 2}$ ways to pick the second pair, etc. so multiplying them would yield all possible first round pairs, but this is not correct and I don't understand why.

Question: Can someone please explain why the above reasoning is not correct


marked as duplicate by Batman, Ross Millikan combinatorics Oct 16 '16 at 21:05

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    $\begingroup$ This is wrong because a vs b, c vs d, e vs f and g vs h is the same as c vs d, a vs b, e vs f and g vs h $\endgroup$ – Astyx Oct 16 '16 at 20:59
  • $\begingroup$ Your way of counting makes difference between, for example, $(a,b)$, $(c,d)$, $(e,f)$, $(g,h)$, but there should be no difference. $\endgroup$ – zar Oct 16 '16 at 20:59
  • $\begingroup$ Your way of choosing pairs overcounts things, in the sense that order matters in it. You should find a way that considers, say, ($a$ vs $b$ and $c$ vs $d$) the same as ($c$ vs $d$ and $a$ vs $b$). $\endgroup$ – Fimpellizieri Oct 16 '16 at 21:00

What you have missed is that a pairing like $(a,b)(c,d)(e,f)(g,h)$ is the same as $(c,d)(e,f)(a,b)(g,h)$ and any of $22$ other such permutations. Your answer would count all $24$ of those separately. So you need to divide your answer by $24$.

  • $\begingroup$ Wow, did not think of that! But this makes a lot of sense. Thanks! $\endgroup$ – ApprenticeOfMathematics Oct 16 '16 at 21:10

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