Answer:
(a) there are 10 sets of 3 friends, so in 21 games, at least one set must show 3 times
(b) there are 20 sets of 3 friends, so in 21 games, at least one set must show 2 times.
Step-by-step explanation:
(a) The number of combinations of 5 things taken 3 at a time is ...
5C3 = 5!/(3!·2!) = 5·4/2 = 10
There can be 10 games in which the same 3 friends do not show up. There can be 10 more games such that the same 3 friends show up exactly twice. In the 21st game, some set of 3 friends must show up 3 times.
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(b) The number of combinations of 6 things taken 3 at a time is ...
6C3 = 6!/(3!·3!) = 6·5·4/(3·2) = 20
Hence, there can be 20 games in which the same 3 friends do not show up. In the 21st game, some set of 3 friends will show up a second time.
