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#101 Re: Help Me ! » Distinguishable objects in indistinguishable bins » 2010-03-22 03:51:43

No, the answer is not so simple.
"abcde||" == "|abcde|" == "||abcde" == "acdeb||" == "ecbad||" == etc...
All permutations of objects inside one bin should be treated as the same combination.
The number of combinations where all objects are inside one of the bins is


When we move one object to another bin we have

When we move two objects to another bin we have

etc...

But how to squeeze all that in one nice formula???

#102 Help Me ! » Distinguishable objects in indistinguishable bins » 2010-03-22 03:31:49

White_Owl
Replies: 13

The Rosen's book, problem 5.5 #50:
How many ways are there to distribute 5 distinguishable objects into 3 indistinguishable bins?

One approach to such problems I know is to imagine distinguishable objects as a long box with compartments, indistinguishable objects as balls and by looking at the balls and separators between compartments we are getting a string of "O" and "l" which can be permuted and that would be the answer...
The only problem here is that as far as I understand this approach, it works perfectly well if number of indistinguishable objects are greater than number of distinguishable.  But if number of distinguishable is greater (like in the problem) I am getting strings like:


a, b, c, d, e here are distinguishable objects represented as compartments. But with this representation I always have some empty compartments - distinguishable objects which are not in any indistinguishable bins...


Another, more promising, approach is to just make a list from letters and separators: "abcde||", "abcd|e|", etc. That would accurately represent distinguishable objects and distinguishable bins. But the original problem states that bins should be indistinguishable, so "abcde||" == "|abcde|" == "||abcde"...
.....hmm....
The process of writing this topic gave me an idea. If we define objects as O and bins as B, then to get the answer it should be enough to count the number of permutations of the string (

) and divide them by number of bins. Am I right or not?

I spent a week thinking about this problem and now I can not believe the answer can be so simple...

#103 Re: Help Me ! » normal distribution » 2010-03-09 02:55:29

Place all your apples in one long line. Starting with the smallest one, ending with the largest one.
Normal distribution means that difference between sizes of two adjusted apples would be same as the difference between any other pair of adjusted apples.

So in your case you have 10% apples which weigh more than 125g and 13% which are less than 75g. That leaves us with 100-10-13=77% apples in the range of 75-125g. Let's assume that in the original batch we had 100 apples (1 apple = 1%).
Equal distribution of 77 good apples through the 100g will give us 1.3g difference in weight between two adjusted apples.
So, to find the smallest apple we can do: 75-13*1.3 = 58.1. The largest apple: 125+10*1.3 = 138.
Once we know the smallest and largest apple in the batch we find the mean: (58.1+138)/2 = 98.05

#104 Re: Help Me ! » simplification of factorials? » 2010-03-03 10:44:23

Oh! I think I got it.
You replaced the

with
because it is the inductive hypothesis...

Arrrr... it is so .... not straight-forward sad

#105 Re: Help Me ! » Surface Area of prisms, im lost :( » 2010-03-03 09:06:43

Yes, the area of the ABC sector is


So the full area of the rounded triangle should be a sum of two sectors minus area of normal equilateral triangle:

Length of the arc BC (or AC):

So the whole surface of this 'curvy triangle tube' should be:

#106 Help Me ! » simplification of factorials? » 2010-03-03 06:58:50

White_Owl
Replies: 4

I am working Rosen's book. Problem 4.1#6:
Prove that


whenever

I am supposed to do it by using induction, so my basis step is:


And the inductive step is (for n+1):

And now I am completely stuck.
Please, point me to the right direction on how to simplify those factorials?

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