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Jakarta June 5th, 2012 06:20 PM

Counting Problem
 
Consider the equation x+y+z+w=n, where n is a positive integer greater or equal to 4. A positive and integer solution is a set (x,y,z,w) of positive integers that satisfies the equation for a given n. For example, for n=10 one solution to x+y+z+w=10 would be (1,2,3,4), since 1+2+3+4=10.
a) Determine the number of positive and integer solutions for n=10.
b) Find a general formula that counts the number of whole and integer solutions for x+y+z+w=n.

Note: Solutions (2, 2, 2, 4), (2, 2, 4, 2),(2, 4, 2, 2) and (4, 2, 2, 2) for n=10 are each considered different and separate solutions.

karthikeyan.jp June 5th, 2012 07:14 PM

Re: Counting Problem
 
so. its finding the no of partitions of n as 4 integers.. it can be split like partitions of n as sum of two numbers.. and finding the partitions of those 2 numbers as sum of two numbers...

karthikeyan.jp June 5th, 2012 10:47 PM

Re: Counting Problem
 
this can be proved by multinomial theorem...

For a+b+c+d = n, the no. of solutions including 0 is



need to eliminate those terms in the expansion of that have powers of all a,b,c,d

johnr June 6th, 2012 03:14 AM

Re: Counting Problem
 
Quote:

Originally Posted by karthikeyan.jp
this can be proved by multinomial theorem...

For a+b+c+d = n, the no. of solutions including 0 is



need to eliminate those terms in the expansion of that have powers of all a,b,c,d

What's the reasoning behind that formula?

karthikeyan.jp June 6th, 2012 05:10 AM

Re: Counting Problem
 
The number of co-efficients in the expansion of is (n+m-1 n). http://en.wikipedia.org/wiki/Multinomial_theorem

So for , it is (n+3 n) ==>

Jakarta June 6th, 2012 05:40 AM

Re: Counting Problem
 
Quote:

Originally Posted by karthikeyan.jp
this can be proved by multinomial theorem...

For a+b+c+d = n, the no. of solutions including 0 is


Well, using this formula, I found through trial and error that the general formula for my problem happens to be:
((n-3)(n-2)(n-1))/6

I'm completely clueless as to the reasoning behind it, though, and politely ask for a baby explanation lol.

soroban June 7th, 2012 03:32 PM

Re: Counting Problem
 
Hello, Jakarta!

Quote:

Consider the equation , where is a positive integer
A solution is a set of positive integers that satisfies the equation for a given

a) Determine the number of solutions for


Place 10 objects in a row, inserting a space between them.
[color=beige]. . [/color]


Select 3 of the 9 spaces and insert "dividers".










Quote:

b) Find a general formula that counts the number of solutions for

Note: Solutions (2, 2, 2, 4), (2, 2, 4, 2), (2, 4, 2, 2) and (4, 2, 2, 2) for
[color=beige]. . . . . [/color]are each considered different and separate solutions.


Following the solution in part (a), place objects in a row (with spaces).

Select 3 of the spaces and insert "dividers".




Jakarta June 7th, 2012 09:05 PM

Re: Counting Problem
 
Quote:

Originally Posted by soroban
Hello, Jakarta!

Quote:

Consider the equation , where is a positive integer
A solution is a set of positive integers that satisfies the equation for a given

a) Determine the number of solutions for


Place 10 objects in a row, inserting a space between them.
[color=beige]. . [/color]


Select 3 of the 9 spaces and insert "dividers".










[quote:3oncju6n]b) Find a general formula that counts the number of solutions for

Note: Solutions (2, 2, 2, 4), (2, 2, 4, 2), (2, 4, 2, 2) and (4, 2, 2, 2) for
[color=beige]. . . . . [/color]are each considered different and separate solutions.


Following the solution in part (a), place objects in a row (with spaces).

Select 3 of the spaces and insert "dividers".



[/quote:3oncju6n]
Wow, that is extremely smart and clear. Thanks!


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