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#1 Re: Help Me ! » 4 points » 2016-03-02 00:40:52

Well, I do not understand the second part with the fractions, but I did it with Excel (generating all possible combinations with repetition) and got the same result smile

MANY THANKS!!

bobbym wrote:

Hi;

There are other ways to attack this problem but from a computational standpoint this is my favorite.

Four points are coplanar iff the determinant of the following matrix is 0.

http://mathworld.wolfram.com/Coplanar.html

Expanding the determinant and setting it to 0 we get:

Dividing by 2400

This is a linear Diophantine equation. The number of solutions to them can be solved by the use of generating functions.

The number of solutions of

with the constraints 
is equal to the coefficient of x^0 ( the constant term ) in the expansion of

This product is nothing more than algebra and can be done by hand methods but it is best done by computer. If you need to I will show you how to take it over to Alpha. This is the power of generating functions, a hard problem has been reduced to a computational one. The expansion of the above product yields:

We see that the constant term is 344 and so the answer is  344 / 8^4 = 43 / 512.

Now I am not saying this is the simplest solution to this problem but these methods have the virtue of being able to solve a great many combinatorics problems with little or no change.

#2 Re: Help Me ! » 4 points » 2016-03-01 12:44:37

Hi bobby; yes, your answer is correct. I would appreciate if you could post your solution.


bobbym wrote:

How?

Before I go into a fairly long and complicated EM method, I would ask if the answer given is correct.

#4 Help Me ! » 4 points » 2016-02-27 05:54:54

kyr_leo
Replies: 8

Find the probability for the following 4 points (0, y1, 0) (60, y2 ,0) (60, y3, 40) (0, y4, 40) to be co-planar, if y1, y2, y3 and y4 can take any integer value from 91 to 98.

#5 Re: Help Me ! » Arrange numbers » 2016-02-24 02:21:35

Agree! What about the numbers 1 to 16?

12
5
14
9
4
7
6
13
16
15
8
3
10
1
2
11

#6 Help Me ! » Arrange numbers » 2016-02-23 22:07:42

kyr_leo
Replies: 3

Can you arrange numbers 1 to 11 on a circle, in such a way that the sum of every two consecutive numbers is a prime number? Is it possible?

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