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dan815
 one year ago
a+b+c+d=k
a,b,c,d>=1
can you find a formula for the number of integer solutions as a function of k?
dan815
 one year ago
a+b+c+d=k a,b,c,d>=1 can you find a formula for the number of integer solutions as a function of k?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It looks like repeated choose operation.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1how did u get that so fast lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2it is a good ol stars and bars problem \(k\) stars and \(3\) bars

dan815
 one year ago
Best ResponseYou've already chosen the best response.1wait wont u have a 0 in the middle with that form

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh did u subtract 3 spots

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for the 3 zeros not allowed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\star\star\star\star\star\star\star\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like if zeros are allowed it would be K+3 choose 3 right

dan815
 one year ago
Best ResponseYou've already chosen the best response.1how come this works, so u subtract 4 spots? for the 4 zeros no allowed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Firstly, notice that those 3 bars divide the 7 stars into 4 pieces

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2You want to have at least 1 star in each piece, yes ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay i see now that limits the total number of places

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats it! so only the places between stars are allowed

dan815
 one year ago
Best ResponseYou've already chosen the best response.1dang that was simple lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i got to this point with some other little trick lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like u know if u look at the difference of the solutions that has a stars and bar solution again allowing 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2solving a+b+c+d = k over nonnegative integers ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1say we were looking at a+b+c=7 then the simple case 1+1+5=7 now i looked at their differences

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can do stars and bars on the differences to see all the possible differences

dan815
 one year ago
Best ResponseYou've already chosen the best response.1only thing is you have to add 1 extra spot in for the differences

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2how do you know there exists a bijection between both the worlds

dan815
 one year ago
Best ResponseYou've already chosen the best response.1well the way i see it is if u write the 1 ,1 ,5 as 1,1,1 +4 now that +4 can be distributed in any way about the 3 terms

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so a+b+c=k can we rewritten as 1+1+(1+k3) and the k3 can be distributed in any way allowing 0 about the 3 terms

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ahh I see so that brings in the evil paritions

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ill post the modified peuler question, i think there are still some cool tricks to be discovered with how to reduce the cases efficiently
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