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pchem

Guest

Need help in probability and statistics

Hi , I cant understand the following :

Can someone please explain this ? I cant understand why we're taking N + g - 1 objects and permuting them.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Need help in probability and statistics

You want to split the n objects into g groups.
If you lay the n objects in a line, that means you need (g-1) dividers.

Let's say there are 5 objects and 3 boxes.
Then one way of distributing them would be 1 in the first box and 2 in the others.
This would correspond to:

.|..|..

By moving the dividers to different positions, you can represent every possible distribution of objects.
The number of arrangements here is 7C2 = 7!/(5!x2!).

Going back to the generalisation of n objects and g groups, the number of arrangements is
(n+g-1)Cn = (n+g-1)!/[n!(g-1)!], as the solution says.

Note that although the objects are indistinguishable, the boxes aren't.
If "All into Box 1" counted as the same as "All into Box 2", then the answer would be different.

---

Why did the vector cross the road?
It wanted to be normal.

pchem

Guest

Re: Need help in probability and statistics

Thanks ... I think I understand better now.

but shouldnt the correct statistical weight be 7C3 = 7!/(4!x3!) ?

Just to set things right , had the objects been distinguishable then the stat weight would be : 
W = (n+g-1)! / [(g-1)!] , correct ?

Had the boxes themselves been indistinguishable do we divide by g! ? i.e.
W = (n+g-1)! / [n!g!(g-1)!] ?

Thanks!

pchem

Guest

Re: Need help in probability and statistics

Suppose we have N objects and we wish to select N1 , N2 , .... Ni objects out of them. The weight would be : W =  N! / N1!N2!...Ni!

But this is as long as N1 objects are indistinguishable. Same thing for  N2 , N3, ... Ni objects.

What if all N objects were indistinguishable ? Wouldnt the weight be W = 1 / N1!N2!...Ni! ?