How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.

Question

The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.

I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.

I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.

Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.

So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.

I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.

Answer 1

Yes, you are absolutely right.

The other approach might be selecting $2$ players out of $6$ since arrangement doesn't matter ($_6C_2=15$).

Answer 2

It's correct (the answer and logic). what is a different way to reason about it?

The idea:

$$(\textrm{first pair})(\textrm{second pair})(\textrm{third pair}),$$

for the first pair: $6$ ways to choose the first guy, $5$ ways to choose the second, and notice the repetition $p_1p_2=p_2p_1$ so $$\frac{6\cdot 5}{2!}$$

for the second and last pair: $\displaystyle\frac{4\cdot3}{2!};\frac{2\cdot1}{2!}.$

Now we get pairs, but $P_1P_2P_3=P_3P_2P_1,$ etc. there are $3!$ ways to repeat$^\dagger$ three pairs we just made, so $$\frac{1}{3!}{6\choose2}{4\choose2}{2\choose2}.$$

$\dagger$: A good question to think about is: what's the condition(s) that repetition will/will not happen?

Answer 3

Your solution is correct.

Here is an alternative approach:

Line up the six players in some order, say alphabetically. The first person in line can choose a partner in $5$ ways. Remove those two players from the line. That leaves four players. The first person left in the line can choose a partner in $3$ ways. Remove those two players from the line. The remaining two players must play each other. Hence, there are $$5!! = 5 \cdot 3 \cdot 1 = 15$$ ways to form three pairs of chess partners. The expression $5!!$ is read $5$ double factorial.