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 November 19th, 2011, 03:07 PM #1 Senior Member     Joined: Feb 2010 Posts: 199 Thanks: 0 How many solutions are there to the equation? ok, this one is really starting to piss me off ... x1 + x2 + x3 + x4 + x5 = 21, x1>=1 i tried all the formulas: n! (n-r)! n!/(r!(n-r)!) (n+r-1)!/(r!(n-1)!) - i think it's this one but nothing's working or even coming close to the answer ... i don't get it how do i do this?
 November 19th, 2011, 03:14 PM #2 Senior Member     Joined: Feb 2010 Posts: 199 Thanks: 0 Re: How many solutions are there to the equation? is it not 25!/(21!*4!) = 12650? the answer is 10626, but that's the closest i can get to it
 November 19th, 2011, 03:18 PM #3 Senior Member     Joined: Feb 2010 Posts: 199 Thanks: 0 Re: How many solutions are there to the equation? 24!/(20!*4!) works tho ... is it because x1 is >=1 and not >=0? ...
 November 19th, 2011, 03:48 PM #4 Senior Member     Joined: Feb 2010 Posts: 199 Thanks: 0 Re: How many solutions are there to the equation? doesn't work far x_i >=2 ... so it's not 23!/(19!*4!) ... the actual answer is 1365
 November 19th, 2011, 04:05 PM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: How many solutions are there to the equation? Let's call f(n, k) the number of solutions to x1 + ... + xn = k with xi >= 0. You're looking for f(5, 16), since this is what you get by assuming each variable has at least 1. It's clear that f(n, k) = f(n-1, k) + f(n-1, k-1) + ... + f(n-1, 0) since this is just saying that xn can have any value from 0 to k. So you have f(5, 16) = f(4, 16) + f(4, 15) + ... + f(4, 0) = f(3, 16) + 2f(3, 15) + 3f(3, 14) + ... + 17f(3, 0) = . . . You can probably solve it this way. You need to keep track of what the coefficients are at each step, but otherwise it's pretty easy. At the end you can simplify f(1, k) = 1 or f(2, k) = k + 1.
November 19th, 2011, 04:14 PM   #6
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Re: How many solutions are there to the equation?

Quote:
 Originally Posted by CRGreathouse You're looking for f(5, 16), since this is what you get by assuming each variable has at least 1. It's clear that f(n, k) = f(n-1, k) + f(n-1, k-1) + ... + f(n-1, 0) since this is just saying that xn can have any value from 0 to k.

could you please explain this a bit more?

November 19th, 2011, 04:15 PM   #7
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Re: How many solutions are there to the equation?

Quote:
 Originally Posted by CRGreathouse So you have f(5, 16) = f(4, 16) + f(4, 15) + ... + f(4, 0) = f(3, 16) + 2f(3, 15) + 3f(3, 14) + ... + 17f(3, 0) = . . . .
also what equation am i supposed to be using for this?

November 19th, 2011, 05:22 PM   #8
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Re: How many solutions are there to the equation?

Quote:
 Originally Posted by Solarmew also what equation am i supposed to be using for this?
The one from my post:
f(n, k) = f(n-1, k) + f(n-1, k-1) + ... + f(n-1, 0)

 November 19th, 2011, 05:33 PM #9 Senior Member     Joined: Feb 2010 Posts: 199 Thanks: 0 Re: How many solutions are there to the equation? ok, i figured than one out and also the one for x_i >=2, i = 1,2,3,4,5 it's 21-5*2 = 11 (11,5) 15!/(11!*4!) but what do i do for 0 =< x1 >= 10?
November 20th, 2011, 04:39 PM   #10
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Re: How many solutions are there to the equation?

Quote:
 Originally Posted by Solarmew ok, i figured than one out and also the one for x_i >=2, i = 1,2,3,4,5 it's 21-5*2 = 11 (11,5) 15!/(11!*4!) but what do i do for 0 =< x1 >= 10?

ahhh... ok... did
364+455+560+680+816+969+1140+133-+1540+1771+2024 = 11649
it works ...

but tried the same approach with

0 =< x1 >= 3
1 =< x4 <4
x3 >= 15

and it didn't work
how come?

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# how many solutions are there to the equation x1 x2 x3 x4 x5=21 where xi,i=1,2,3,4,5,is a non negative integer such that x1>=1?

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