數學嘅難題｛１５點｝

我地要個條式要包含哂咁多個答案，而且係有８個人９個人個陣都用得......

１條式過的組合計法:
nCr=n!/r!/(n-r)!

n=總人數 eg. 7人
r=多少人一組 eg.4人.....
r!,n!=乘階 eg.7!=1*2*3*4*5*6*7=5040
eg. 4!=1*2*3*4=24

例:依家有７個人,四個人一組
nCr=n!/r!/(n-r)!
=7!/4!/(7-4)!
=5040/24/6
=35組
例:依家有9個人,四個人一組
nCr=n!/r!/(n-r)!
=9!/4!/(9-4)!
=362880/24/120
=126組

依家有７個人，我地要幫佢分組，二個人一組，三個人一組，四個人一組，五個人一組，六個人一組

我地唔係要個答案，我只係要個式，好似一個正方形佢個式係４×４＝１６咁（邊長係４）我地係要條式（好似岩岩個正方形嘅４×４咁）不過要１條式過，唔可以係好多條式，仲要有解釋

Ok, it was pretty hard to understand your question, but I think I got it.

Let A B C D E F G be the 7 people, you question is how many groups can be formed by these person given there are n number of people in the group.

If n is 2, then you have:

AB, AC, AD, AE, AF, AG, BC, BD...to EF, EG, FG

and there are 21 unique groups can be formed from this.

Let say we choose any one of these people into a group.

We have 7 people here, so we have 7 choices for the first member of this group.

For the second member of this group, because one of the people on the list has been picked out, there are 6 choices left.

So in total there are 7x6 OR 42 choices to pick our groups. BUT some of these arrangements are the same. For example, we have

AB, AC, AD, AE, AF, AG,
BA, BC, BD, BE, BF, BG.....

AB and BA are two arragements that has the same people in it.

In order to elimate the redundant arragements from our result, we need to divide it by the number of ways this arragement can be differed.

That is, in the case of AB and BA, there are only 2 different ways to rearrage this, same applied by BC and CB and others.

So, we have to divide it by our result by 2.

That is: in total there are 7x6/2 OR 21 unique groups if the groups are made up of 2 people.

2008-07-23 19:18:36 補充：
The explanation is a bit long, the rest is saved on this link:
http://img155.imageshack.us/img155/9845/explainationvv2.png