[理工] 排列組合
A father buys nine different toys for his four children.
In how many ways can he give one child three toys and the
remaining three children two toys each?
我的想法是先分堆再給物品(3,2,2,2)因為四個小孩都有可能
得到3個玩具所以我在乘上4!/3!=4,得到
4*(9C3)*(6C2)*(4C2)*(2C2)=30240
但答案卻是9!/3!2!2!2!=7560,為什麼他沒考慮
不同人取得三玩具之情形?
感謝回答
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.193.7.20
推
10/14 21:48, , 1F
10/14 21:48, 1F
→
10/14 21:56, , 2F
10/14 21:56, 2F
→
10/14 21:56, , 3F
10/14 21:56, 3F
→
10/14 22:26, , 4F
10/14 22:26, 4F
討論串 (同標題文章)
