Math Is Fun Forum

So my formula may work by adding some error correction function:

And, going to the classical way:
We have 0 combinations with 1 balls.
b combinations with 2 balls (each box filled with 2 balls and the others - empty)
b(b-1) combinations with 3 balls (filling some box with 2 balls and other with one)
b(b-1)(b-2) combinations with 4 balls (filling as 3 balls and another ball)
...
b(b-1)(b-2)...(b-k+2) = b!/(b-k+1)! combinations with k balls

By Pigenhole principe, we should stop at k=b+1.
Now, just finding AM:

A program:

#include<iostream>
#include<math.h>
using namespace std;
int main(){
	const int MAXBINS=100;
	long bins, tests, btmp[MAXBINS], current, counter, res, i, j, rnd;
START:
	cout<<"Please enter the number of bins:";
	cin>>bins;
	cout<<"Please enter the number of tests:";
	cin>>tests;
	res=0;
	for(i=0;i<tests;i++){
		counter=0;
		for(j=0;j<bins;j++)
			btmp[j]=0;
		while(1){
			rnd = floor((double)rand()*bins/RAND_MAX);
			btmp[rnd]++;
			counter++;
			if(btmp[rnd]==2) break;
		}
		res+=counter;
	}
	cout<<"The result is:"<<(double)res/tests<<endl;
	cout<<"Press 1 for new test, 0 for exit:";
	cin>>i;
	if(i!=0) goto START;
	return 0;
}

mathsy's right.
With horner's scheme (i don't know exactly how's in English): ( changing the sign:)
   | 2  -5   2     5
2 | 2  -9  20  -35 ->remainder;

Just an obsrvation - because the numbers are in base 2, the total amount of the numbers, starting with 1 is equal to the total amount of numbers, starting with 0. But we can't use this in this problem, because the number of the nulls is fixed.

I've founded a quick way to calculate this!
We have 2 A-s, 2 E-s and 2 T-s.
Let us index the same letters, e.a. A1,A2,E1,E2,T1,T2.
Now we'll use the system{A1,A2,E1,E2,T1,T2,L,R,N}.
In this system the words
A1 LTERN A2 TE
and
A2 LTERN A1 TE
are different, but in the original system, they are the same word.
Now, let's meashure the following:
1 word form the normal system = ? words form the extended system?
Any "normal" word won't change, if we substitute A1 with A2 and E1 with E2 and T1 with T2. 3 couples in total , so:
1 word form the normal system =2^3=8 words form the extended system.
Now, we just need to find the word form the extended system and divide by 8.
In the extended systen, we have no couples, at the first place should stand 1 of 4 letters - A1, A2, E1 or E2. at the last place we should place one of the left 3. Between them, we'll have 7 "empty slots" for filling, which could be done in 7! ways. So the answer is:

What are these calculations?
And the gravity at poles is different than the gravity at the Equatior.

But in some cases, an accurate result is needed.
er, could you tell us what do you need it for?

He he!
Joke, right?

//just a test...

This remids me about something...
Since:

That was a tricky one.
It gave me inspiration,too.
Try with the word PENCILLS.
I can rephrase the Stanley's solution:
Let S be the set of all words.
We have 2 sets of words:
A)..xxNx..xLxx..
B)..xxLx..xNxx..
Then A U B = S and A . B = {} (here "." stands for 'or')
And, there exists a bijection θ: A->B, so |A|=|B|.
We can use simular observations to prove the following theorem:
Let