Question

Eight friends want to play enough games of chess to be sure everyone plays everyone else. How many games will they have to play?

Answer 1

Everyone will have to play 21 games

Answer 2

The number of games, the friends have to play is an illustration of combination.

They have to play 28 games to ensure that they all play

Given

$n = 8$ --- number of friends

$r = 2$ --- only two friends can play at a time

For everyone else to play, we will make use of the following combination formula;

$^nC_r = \frac{n!}{(n - r)!r!}$

So, we have:

$^8C_2 = \frac{8!}{(8 - 2)!2!}$

$^8C_2 = \frac{8!}{6!2!}$

Expand

$^8C_2 = \frac{8 \times 7 \times 6!}{6! \times 2 \times 1}$

$^8C_2 = \frac{8 \times 7}{2 \times 1}$

$^8C_2 = 4 \times 7$

$^8C_2 = 28$

Hence, the 8 friends have to play 28 games to ensure that everyone else plays

Read more about combinations at:

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