+63 Our basketball team has 10 players, including Steve and Danny. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this if Steve and Danny insist on playing on the same team?
Help! BTW the answer is not 126.
+6251 \(\text{Steven and Danny are on the same team so there are 8 players left to choose from}\\ \text{The number of ways teams can be arranged is}\\ \dbinom{8}{3}=\dbinom{8}{5} = 56\\ \text{i.e. you can look at it as choosing the 3 players for the team they are on,}\\ \text{or choosing the 5 of the team they aren't. They are equivalent}\)
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