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Ten basketball players want to divide themselves into $2$ teams of $5$ players each,in such a way that the $2$ best players are on the opposite teams. In how many ways can this be done?

I have two potential answers in mind. One is $\frac{\binom{8}{4}}{2}$ and the other is $\frac{\binom{8}{4}}{2}$ . I find it hard to decide from which total number of people should be chosen from.

Thanks in advance!

user235684
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JimfyWinsy
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  • The first one is correct: we separate 2 "best players", mention one of them and choose 4 of 8 of the rest to his team and the rest 4 to another. – Alexey Burdin May 02 '15 at 05:14

3 Answers3

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Whoever the best $2$ players are, we let the one with the lower student number pick her team mates. She can do this in $\binom{8}{4}$ ways.

André Nicolas
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You have to consider one player in the first team (say A) to be static and fill the other 4 positions with 8 players, then you get 8C4 combinations. This automatically calculates who will be on the other team. since the two best players can be interchanged among the teams you have to multiply by 2. Answer comes 2x8C4

slhulk
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  • Your answer is incorrect. Choosing four players to be on the team of the best player determines who will be on the team of the second best player, so it suffices to choose the teammates of the best player. – N. F. Taussig May 02 '15 at 10:29
  • I agree. Then 2x 8C4 only – slhulk May 02 '15 at 10:42
  • Your answer is still incorrect. By switching the two best players, you have counted each team twice. Let A, B, C, D, E, F, G, H be the players to be selected. For instance, you count the team consisting of the best player and players A, B, C, and D once when you select them to be on the best player's team and once when you first select players E, F, G, and H to be on the best player's team, then switch the two best players. – N. F. Taussig May 02 '15 at 12:39
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I'd ask those two best (suppose 1,2) players to stand on opposite sides of the court. Now I'd choose 4 from the remaining 8. So the answer is (8C4).

GrandAlpha
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