.(x+y+z+u)(7次方)展開式中,不同的項共有幾項

2008-01-24 9:23 pm
.(x+y+z+u)(7次方)展開式中,不同的項共有幾項
更新1:

thanks for your answer, but I don't want to get the expanssion I want to get the way for "how to calculate 不同的項共有幾項?" can you show me? thanks

回答 (1)

2008-01-24 10:49 pm
✔ 最佳答案
In each bracket, we can only choose 1 item.
As each term is degree 7, we need to choose 7 items from 7 bracket respectively.
It is the same as choosing 7 items from 4 items with replacement, i.e. 4H7.
4H7
=(4+7-1)C7
=10C7
=10C3
=120


收錄日期: 2021-05-01 18:35:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080124000051KK01044

檢視 Wayback Machine 備份