You are not logged in.
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
But how to squeeze all that in one nice formula???
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:
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 (
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
Oh! I think I got it.
You replaced the
Arrrr... it is so .... not straight-forward
Yes, the area of the ABC sector is
Length of the arc BC (or AC):
So the whole surface of this 'curvy triangle tube' should be:
I am working Rosen's book. Problem 4.1#6:
Prove that
I am supposed to do it by using induction, so my basis step is: