v********w
发帖数: 136 | | p******5
发帖数: 138 | 2 If there are i ranks for a final result, then the number of different
results is C(7, i-1), that is, the number of the different way of picking i-
1 elements from the set of 7 elements.
So the total number of results is \sum^8_{i = 1} C(7, i-1). | l*******1
发帖数: 113 | 3 what kind of ties do you allow?
for abcdefgh
can u have random ties like ab, cd, efg, h? and abc, de, fgh? | G********d
发帖数: 10250 | 4 in your computation
ab cd efg
cd ab efg
are only counted once.
If there are i ranks for a final result, then the number of different
results is C(7, i-1), that is, the number of the different way of picking i-
1 elements from the set of 7 elements.
So the total number of results is \sum^8_{i = 1} C(7, i-1).
【在 p******5 的大作中提到】
: If there are i ranks for a final result, then the number of different
: results is C(7, i-1), that is, the number of the different way of picking i-
: 1 elements from the set of 7 elements.
: So the total number of results is \sum^8_{i = 1} C(7, i-1).
| p******5
发帖数: 138 | 5 You are right.
How about \sum^8_{i = 1} C(7, i-1)*i! ?
picking i-
【在 G********d 的大作中提到】
: in your computation
: ab cd efg
: cd ab efg
: are only counted once.
:
: If there are i ranks for a final result, then the number of different
: results is C(7, i-1), that is, the number of the different way of picking i-
: 1 elements from the set of 7 elements.
: So the total number of results is \sum^8_{i = 1} C(7, i-1).
| G********d
发帖数: 10250 | 6 then
ab cd efg
ba cd efg
are both counted
my answer
\sum_{n=1}^8 \sum_{i=0}^{n-1} (n-i)^8(-1)^i n!/(i!(n-i)!)
You are right.
How about \sum^8_{i = 1} C(7, i-1)*i! ?
picking i-
【在 p******5 的大作中提到】
: You are right.
: How about \sum^8_{i = 1} C(7, i-1)*i! ?
:
: picking i-
| b******n
发帖数: 637 | | v********w
发帖数: 136 | 8 说说看？
楼上的能不能解释一下你的式子含义？
【在 b******n 的大作中提到】
: 有表达式，但是求不出close form...
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