h*********g
发帖数: 380 | 1 How many ways are there to distribute 20 different toys among five children
with each child should get at least one?
Thanks very very much! |
H****h
发帖数: 1037 | 2 用容斥原理来计算。
children
【在 h*********g 的大作中提到】
: How many ways are there to distribute 20 different toys among five children
: with each child should get at least one?
: Thanks very very much!
|
h*********g
发帖数: 380 | 3 Could you please give more details? Thanks
【在 H****h 的大作中提到】
: 用容斥原理来计算。
:
: children
|
k*****l
发帖数: 177 | 4
0|00|00000|000|000000000
Use four walls to split above string...
C(19,4)*20!
children
【在 h*********g 的大作中提到】
: How many ways are there to distribute 20 different toys among five children
: with each child should get at least one?
: Thanks very very much!
|
h*********g
发帖数: 380 | 5 C(19,4) has no problem, but then multiple 20! will have many overlapping
results. We can not just simply arrangement 20 toys, cause arrangement of
different toys when they are all assigned to one kid has no meaning.
【在 k*****l 的大作中提到】
:
: 0|00|00000|000|000000000
: Use four walls to split above string...
: C(19,4)*20!
:
: children
|
w********9
发帖数: 8613 | 6 take any 5 toys out of the 20：
20！/ 5! / 15!
allocate the 5, one for each child:
5!
allocate the remaining 15:
5^15
multilying all of the above:
5^15 * 20! / 15! |
H****h
发帖数: 1037 | 7 定义A1是第一个小朋友没有拿到玩具的所有可能组成的集合。类似有A2至A5。
你要计算的是所有A集合之补集相交得到的集合的元素个数。
【在 h*********g 的大作中提到】
: Could you please give more details? Thanks
|
l******r
发帖数: 48 | 8 Happyeating shit!
You are a mother f**ker!
Are you looking for a good algorithm to f**k your mother?? |
