[問題]天山04-18

[feebs79]

Department A, B, and C have 10 employees each, and department D has 20 employees. Department A, B, C, and D have no employees in common. A task force is to be formed by selecting 1 employee from each of the departments A, B, and C and 2 employees from department D. How many different task forces are possible?

A)19,000
B)40,000
C)100,000
D)190,000
E)400,000

Answer: D

但是我算法是C(10 , 1)* C(10 , 1)*C(10 , 1)*C(20 , 2)=380,000

請問我哪裡出錯了?

謝謝

[zbow]

Department A, B, and C have 10 employees each, and department D has 20 employees. Department A, B, C, and D have no employees in common. A task force is to be formed by selecting 1 employee from each of the departments A, B, and C and 2 employees from department D. How many different task forces are possible?

A)19,000
B)40,000
C)100,000
D)190,000
E)400,000

Answer: D

但是我算法是C(10 , 1)* C(10 , 1)*C(10 , 1)*C(20 , 2)=380,000

請問我哪裡出錯了?

謝謝

Ｃ(20,2) = 190 ...　

應該是計算錯了吧. ^^

[feebs79]

C(20,2)=20!/(20-2)!=20!/18!=20x19=380
不是嗎?

還是...是我...終於瘋了!!

[zbow]

:P ...
C(20,2) = 20! / (2!(20-2)!))

看書看太累了...可能要休息一下 ... XD