In a tournament, each of the $6$ participants played 2 matches against each of the other participants. What was the total number of matches played during the tournament?
So we have a set of 6 participants.
$A, B, C, D, E, F$
So $A$ can play with either $B, C, D, E, F$ and likewise for the other participants.
$A$ has $5$ options, $A$ can choose $2$ options.
But, how can we solve this? I can just base the layout, nothing else =(
On
I am reposting my revised comment.
A plays 2 games with each of the other five players $= 5*2$
B plays 2 games with each of the other four players, (A needs to be discounted this time because A playing B is the same as B playing A) $= 4*2$
Thus the argument extends to rest of the cases. ... Goes on until
E plays 2 games with one player $= 1*2$
Add them all : $2\times(5+4+3+2+1)= 6*5 = 30$.
In a combinatorial way it is ${6\choose2}\times 2 = 30$
First let's say "naievely": each of the $6$ participants plays $5\times 2=10$ matches, so there will be $6\times 10=60$ matches in total. Then realize that the match $A$ against $B$ is counted twice this way: as a match played by $A$ and as a match played by $B$. This is true for each match. We can repair this by dividing the number we found on the naive way by $2$ resulting in a total of $30$ matches.