Question

There are $5$ girls and $5$ boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once.

In how many games does a girl play against another girl?

Answer 1

10

Explanation:

Answer 2

Let's call the five girls
`g_1, g_2, g_3, g_4 \text {and } g_5`.

Since every player plays against every other player exactly once,
`g_1` has to play against
`g_2, g_3, g_4 \text {and } g_5`. This are four maches.

`g_2` has already played against
`g_1`, so she has to play against
`g_3, g_4 \text {and } g_5`. This are three matches.

`g_3` has already played against
`g_1 \text{ and } g_2`, so she has to play against
`g_4 \text {and } g_5`. This are two matches.

`g_4` has already played against
`g_1, g_2 \text{ and } g_3`, so she has to play against
`g_5`. This is one match.

So, the total is
`4+3+2+1 = 10` matches.