Question

There are 27 players in a ping-pong tournament. each player plays every other player exactly once. how many games will be played?

Answer 1

if x people play each other in pairs exactly once, then there will be
hmm

if 1, then 0 games
if 2, then 1 game
if 3 then 3 games ab, ac, bc
if 4 then ab, ac, ad, bc, bd, cd then 6 games
if 5 then ab, ac, ad, ae, bc, bd, be, cd, ce, de
we see a pattern
summation
first term is 0, when n=1
increases by 1
so
$S_n=\frac{n(0+n-1)}{2}$ simlified to
$S_n=\frac{n(n-1)}{2}$

for n games, a total of $\frac{n(n-1)}{2}$ games are played

for 27 players, there are a total of $\frac{27(27-1)}{2}=\frac{27(26)}{2}=$ 27(13)=351 games