combination math question

I see the answer is 12c4* 8c4*1, why not times 6 since the coach can also be different for different team?

Thomas, Carrie, and Lenny each captain a different one of three hockey teams. Each captain will choose four players from a pool of 12 players, with each player chosen for only one team. How many different ways can the teams be formed?

再乘就重复了阿。 先固定教练， 第一个教练 12 选 4 ， 第二个教练只能 8 选 4， 剩下的归第三个教练。

assume 3 players instead of 12. write it down and you'll see each coach already got assigned to all combos so that you can't x6. why? when you grab a coin on each hand, with 2 different coins you only have two ways - you can't x2.

I see the answer is 12c4* 8c4*1, why not times 6 since the coach can also be different for different team?

Thomas, Carrie, and Lenny each captain a different one of three hockey teams. Each captain will choose four players from a pool of 12 players, with each player chosen for only one team. How many different ways can the teams be formed?
ll2eq1 发表于 2025-01-17 15:24

你把12个人里面选4个变成3个人里面选一个你就明白了
如果三个人选一个答案应该是3 choose 1 times 2 choose 1 times 1 choose 1
对应的就是 a1 b2 c3 (i) a1 b3 c2 (ii) a2 b1 c3 (iii) a2 b3 c1 (iv) a3 b1 c2 (v) a3 b2 c1 (vi)
如果你这里再乘以6的话等于说再把 abc这里再做permutation
比如说你把abc变成了bca 那么 (i) a1 b2 c3 -> b1 c2 a3 其实是和 (v) 重复的

ll2eq1 发表于 2025-01-17 15:24
I see the answer is 12c4* 8c4*1, why not times 6 since the coach can also be different for different team?

Thomas, Carrie, and Lenny each captain a different one of three hockey teams. Each captain will choose four players from a pool of 12 players, with each player chosen for only one team. How many different ways can the teams be formed?

duplicate

combined

The choosing order does not matter here, only who the captains choose matter. If order matters you could use permutation that is times 6. For example, T choose batch x first, C choose batch y second, L choose batch z third, if the problem considers the choosing order, then a different answer is C choose batch y first, T choose batch x second, L choose batch z last. But in your problem, which captain choose first does not matter as long as they have the same team attribution as final result. In this case, we only use combination.