I need to find the total number of ways to choose 5 players out of 12, and then subtract the number of ways that both Bob and Yogi are chosen. To choose 5 players out of 12, I can use the formula for combinations, which is 12 choose 5, or 12!/(5!)(12-5)!, which is equal to 792. To choose both Bob and Yogi, I need to choose the other 3 players out of the remaining 10 players, which is 10 choose 3, or 10!/(3!)(10-3)!, which is equal to 120. So, the number of ways to choose 5 players out of 12 that do not include both Bob and Yogi is 792 - 120, which is 672.


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.)