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It looks like repeated choose operation.
how did u get that so fast lol
it is a good ol stars and bars problem \(k\) stars and \(3\) bars
wait wont u have a 0 in the middle with that form
oh did u subtract 3 spots
for the 3 zeros not allowed
like if zeros are allowed it would be K+3 choose 3 right
how come this works, so u subtract 4 spots? for the 4 zeros no allowed
Firstly, notice that those 3 bars divide the 7 stars into 4 pieces
You want to have at least 1 star in each piece, yes ?
okay i see now that limits the total number of places
thats it! so only the places between stars are allowed
dang that was simple lol
i got to this point with some other little trick lol
like u know if u look at the difference of the solutions that has a stars and bar solution again allowing 0
solving a+b+c+d = k over nonnegative integers ?
kinds cool that exists
yeah like umm
say we were looking at a+b+c=7 then the simple case 1+1+5=7 now i looked at their differences
you can do stars and bars on the differences to see all the possible differences
only thing is you have to add 1 extra spot in for the differences
how do you know there exists a bijection between both the worlds
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
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
Ahh I see so that brings in the evil paritions
ill post the modified peuler question, i think there are still some cool tricks to be discovered with how to reduce the cases efficiently