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Problem : Suppose there are ${k}$ teams playing a round robin tournament; that is, each team plays against all the other teams and no game ends in a draw.Suppose the ${i^{th}}$ team loses ${l_{i}}$ games and wins ${w_{i}}$ games.Show that

${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$ = ${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$

Solution : Each team plays exactly one match against each other team.

Consider the expression $\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) }$

Since each team plays exactly k-1 matches and no match ends in a draw, hence number of wins plus numbers of loses of a particular team is k-1 (that is the number of matches it has played). In other words $l_i + w_i = k-1$ for all i (from 1 to k).

Hence

$\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} }$
$\displaystyle{= \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) }$
$\displaystyle{= \sum_{i=1}^{k}(k-1)(l_i - w_i) }$
$\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) }$

But $\displaystyle{ \sum_{i=1}^{k} l_i = \sum_{i=1}^{k} w_i }$ (as total number of loses = total number of matches = total number of wins; as each match results in a win or lose of some one)

Hence $\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) = (k-1) \times 0 = 0 }$

Therefore $\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = 0 }$ implying ${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$ = ${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$

Proved.