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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
MANY THANKS!!
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 ofThis 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.
Hi bobby; yes, your answer is correct. I would appreciate if you could post your solution.
How?
Before I go into a fairly long and complicated EM method, I would ask if the answer given is correct.
How?
Hi
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.
Agree! What about the numbers 1 to 16?
12
5
14
9
4
7
6
13
16
15
8
3
10
1
2
11
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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