dan815
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
It looks like repeated choose operation.
ganeshie8
  • ganeshie8
\[\binom{k-1}{3}\]
dan815
  • dan815
how did u get that so fast lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
it is a good ol stars and bars problem \(k\) stars and \(3\) bars
dan815
  • dan815
wait wont u have a 0 in the middle with that form
dan815
  • dan815
oh did u subtract 3 spots
dan815
  • dan815
for the 3 zeros not allowed
ganeshie8
  • ganeshie8
\[\star\star|\star\star|\star|\star\star\]
dan815
  • dan815
like if zeros are allowed it would be K+3 choose 3 right
ganeshie8
  • ganeshie8
Yep
dan815
  • dan815
how come this works, so u subtract 4 spots? for the 4 zeros no allowed
ganeshie8
  • ganeshie8
Firstly, notice that those 3 bars divide the 7 stars into 4 pieces
dan815
  • dan815
right
ganeshie8
  • ganeshie8
You want to have at least 1 star in each piece, yes ?
dan815
  • dan815
right
dan815
  • dan815
okay i see now that limits the total number of places
dan815
  • dan815
ahhh
ganeshie8
  • ganeshie8
thats it! so only the places between stars are allowed
dan815
  • dan815
dang that was simple lol
dan815
  • dan815
i got to this point with some other little trick lol
dan815
  • dan815
like u know if u look at the difference of the solutions that has a stars and bar solution again allowing 0
ganeshie8
  • ganeshie8
solving a+b+c+d = k over nonnegative integers ?
dan815
  • dan815
kinds cool that exists
dan815
  • dan815
yeah like umm
dan815
  • 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
  • dan815
|dw:1435697202677:dw|
dan815
  • dan815
you can do stars and bars on the differences to see all the possible differences
dan815
  • dan815
only thing is you have to add 1 extra spot in for the differences
ganeshie8
  • ganeshie8
how do you know there exists a bijection between both the worlds
dan815
  • 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
  • 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
  • ganeshie8
Ahh I see so that brings in the evil paritions
dan815
  • dan815
yaa
dan815
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.