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

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

It looks like repeated choose operation.

- ganeshie8

\[\binom{k-1}{3}\]

- dan815

how did u get that so fast lol

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## More answers

- ganeshie8

it is a good ol stars and bars problem
\(k\) stars and \(3\) bars

- dan815

wait wont u have a 0 in the middle with that form

- dan815

oh did u subtract 3 spots

- dan815

for the 3 zeros not allowed

- ganeshie8

\[\star\star|\star\star|\star|\star\star\]

- dan815

like if zeros are allowed it would be K+3 choose 3 right

- ganeshie8

Yep

- dan815

how come this works, so u subtract 4 spots? for the 4 zeros no allowed

- ganeshie8

Firstly, notice that those 3 bars divide the 7 stars into 4 pieces

- dan815

right

- ganeshie8

You want to have at least 1 star in each piece, yes ?

- dan815

right

- dan815

okay i see now that limits the total number of places

- dan815

ahhh

- ganeshie8

thats it!
so only the places between stars are allowed

- dan815

dang that was simple lol

- dan815

i got to this point with some other little trick lol

- dan815

like u know if u look at the difference of the solutions that has a stars and bar solution again allowing 0

- ganeshie8

solving a+b+c+d = k over nonnegative integers ?

- dan815

kinds cool that exists

- dan815

yeah like umm

- dan815

say we were looking at a+b+c=7
then the simple case
1+1+5=7
now i looked at their differences

- dan815

|dw:1435697202677:dw|

- dan815

you can do stars and bars on the differences to see all the possible differences

- dan815

only thing is you have to add 1 extra spot in for the differences

- ganeshie8

how do you know there exists a bijection between both the worlds

- dan815

well 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

so a+b+c=k
can we rewritten as
1+1+(1+k-3)
and the k-3 can be distributed in any way allowing 0 about the 3 terms

- ganeshie8

Ahh I see so that brings in the evil paritions

- dan815

yaa

- dan815

ill 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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