# How many ways to pair tennis players? [duplicate]

I've come across the following problem in my textbook

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 16 '16 at 21:05

• Your way of counting makes difference between, for example, $(a,b)$, $(c,d)$, $(e,f)$, $(g,h)$, but there should be no difference. – zar Oct 16 '16 at 20:59
• 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 $b$). – 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 $24$.