Question 280532
each team plays with each other twice.
there are a total of 30 games played.
how many teams?


If each team played with each other once, the number of games played would have been 15.


For each team to play with each other once, the following formula applies:


{{{15 = (x!)/(((x-2)!)*(2!))}}}


This is a combination formula where x is the number of teams and 2 is the number of teams required to play each game.


The questions being asked is what value of x allows us to get 15 sets of 2 where each set is unique.


The only way I know how to solve this is by trial and error.


I tried a few numbers until I got to 6 which is the right number.


When x = 6, the formula becomes:



{{{15 = (6!)/(((6-2)!)*(2!))}}} which becomes:



{{{15 = (6!)/((4!)*(2!))}}} which becomes:


{{{15 = (6*5*4*3*2*1)/((4*3*2*1)*(2*1))}}} which becomes:


{{{15 = (6*5)/(2) = 30/2 = 15}}}


Since 15 = 15, the value of x = 6 is good.


We have 15 games where 6 teams play with each other once.


If they play with each other twice, the number of games is equal to 30.


We can see how this works out if we assign each team a letter.


The letters are abcdef


The possible ways for 6 teams to play with each other once is:


ab
ac
ad
ae
af
bc
bd
be
bf
cd
ce
cf
de
df
ef


That's 15 games if they play with each other once, 30 games if they play with each other twice, 45 games if they play with each other 3 times, etc.