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counting and probability intermediate
Problem
The Screamers are coached by Coach Yellsalot. The Screamers have 12 players, but two of them, Bob and Yogi, refuse to play together. How many starting lineups (of 5 players) can Coach Yellsalot make, if the starting lineup can't contain both Bob and Yogi? (The order of the 5 players in the lineup does not matter; that is, two lineups are the same if they consist of the same 5 players.)
Solution
There are 3 different cases for the starting lineup.
Case 1: Bob starts (and Yogi doesn't). In this case, the coach must choose 4 more players from the 10 remaining players (remember that Yogi won't play, so there are only 10 players left to select from). Thus there are lineups that the coach can choose.
Case 2: Yogi starts (and Bob doesn't). As in Case 1, the coach must choose 4 more players from the 10 remaining players. So there are lineups in this case.
Case 3: Neither Bob nor Yogi starts. In this case, the coach must choose all 5 players in the lineup from the 10 remaining players. Hence there are lineups in this case. To get the total number of starting lineups, we add the number of lineups in each of the cases:
Case 1: Bob starts (and Yogi doesn't). In this case, the coach must choose 4 more players from the 10 remaining players (remember that Yogi won't play, so there are only 10 players left to select from). Thus there are lineups that the coach can choose.
Case 2: Yogi starts (and Bob doesn't). As in Case 1, the coach must choose 4 more players from the 10 remaining players. So there are lineups in this case.
Case 3: Neither Bob nor Yogi starts. In this case, the coach must choose all 5 players in the lineup from the 10 remaining players. Hence there are lineups in this case. To get the total number of starting lineups, we add the number of lineups in each of the cases:
Final answer
672