**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.