There are 10 people playing basketball. We want to divide them into two teams of 5. However, Dan and Mike want to be on the same team. How many combinations are there where Dan and Mike are on the same team?
2026-03-25 17:44:40.1774460680
There are 10 people playing basketball. We want to divide them into two teams of 5. However, Dan and Mike want to be on the same team.
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Place Dan and Mike on the same team. There are $8$ remaining players. There are $$\binom{8}{3}$$ ways to choose the $3$ teammates for Dan and Mike, and the remaining $5$ players are on the other team.
This means that the answer is simply $$\binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2} = \boxed{56\,}$$