## 2008年2月27日 星期三

### A Quote

Suppose that there is a picnic consisting of many families and we want to count the number of families. One way would be to define some "canonical head" of each family, say "mother", and count the number of mothers. But some daughters look like mothers, so this is not so easy. On the other hand, you cannot just count everybody , since then you would count each family several times. The problem is that "naive" counting of people is giving a credit of 1 to each person, and this is inappropriate if we are trying to count families. If instead we were to ask each person "How big is your family?" and add to our count the reciprocal of that number, then the calculation would come out just right, since a family of size k would get a credit of 1/k for each of its members, and would therefore have been counted exactly once by the end.

Going back to counting orbits, we see by the same reasoning that their number is \sum_{a \in A} 1 / | Orbit(a) | ......

#### 3 則留言:

Kahoo 提到...

a_n = sum(k=0 to [n/2]) C(n-k,k)

(*)　　C(n,k) + C(n,k+1) = C(n+1,k+1)

Marco_Dick 提到...

A rabbit can jump up one stair or two stairs each time. The number of ways to jump to the n-th stairs is exactly a_n. Another way to count it can be considering n-k as the number of jumps,

Kahoo 提到...

Wow, very nice!