Nineteen students are eligible to play doubles tennis. What is the maximum number of different 2-person teams possible?

Summary Submit Skip Question

This is a combination since order does not matter (for groups of two).

$$ {}_nC_r = \frac{n!}{(n-r)!r!} $$ $$ {}_{19}C_2 = \frac{19!}{(19-2)!2!} $$ $$ = \frac{19!}{17!2!} $$ $$ = \frac{19\cdot 18}{2} $$ $$ = 19\cdot 9 = \boxed{171} $$

The first player can pair up with any of the other 18 players.

The second player can pair up with any of the 18 other players, but pairing up with the first player would be a duplicate pairing. So there are 17 options for the second player.

This trend continues:

$$ 18, 17, 16, 15 ... 1 $$

We can sum the arithmetic sequence manually, or with the formula:

$$ s_n = \frac{n(n+1)}{2} $$ $$ s_{19} = \frac{18(18+1)}{2} $$ $$ s_{19} = 9\cdot 19 = \boxed{171} $$