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# Combinatorics

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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.

Feb 15, 2019

$$\text{Steven and Danny are on the same team so there are 8 players left to choose from}\\ \text{The number of ways teams can be arranged is}\\ \dbinom{8}{3}=\dbinom{8}{5} = 56\\ \text{i.e. you can look at it as choosing the 3 players for the team they are on,}\\ \text{or choosing the 5 of the team they aren't. They are equivalent}$$