/sci/ - Science & Math

Will keep bumping until /sci/ stops posting in 9/11 conspiracy threads and debates about religion.

You sure you wrote the problem down correctly?
p^(k choose 2) is constant. And (n choose k) will grow as n -> infinity.

>>5073542
I though that too at first, but by the "=1 condition", d and n (or k and n) are not independend. I think.

Here are some ideas for a pedestrian version, I didn't get far thoug:

We investigate
(n choose k)*p^(k choose 2)
and for fixed k, this would of course blow up.
Now appearently this seems not to happen, and that means that the only real difficult part seems to be the "=1" condition, i.e. the dependence of n on k (or resp. d).

For starters, let's set

k=d+x
with
x=1+y,
with
y>=0
as the problem says.
(For y=0 you have the k=d+1, for y>0 you have k>d+1)

z choose 2 = -z/2 + z^2/2

so

(d+x) choose 2
= -d/2 + d^2/2 + x*(x-1+2d)/2
= (d choose 2) + c

with
c := x*(x-1+2d)/2 >0
which is > 0, because x>1 (assuming d>0)

p^(k choose 2)
=p^((d+x) choose 2)
=p^(d choose 2) * p^c (now using condition "=1")
=(n choose d)^-1 * p^c
=(n choose d)^-1 * C

with some number 0<C<1

so

(n choose k) * p^(k choose 2)
=(n choose (d+x)) * p^(k choose 2)
=(n choose (d+x))/(n choose d) * C

...?

fuck you combinatorics

>>5073542
k isn't constant. d depends on n, and k is greater than d.

Obviously the thing equal to 1 has to be used in some way... can't for the life of me see how, though.

Sorry, I don't bother helping people who are too stupid and lazy to latex their shit.

>>5073674
> "I don't talk to stupid people"
> Can't do a clear and simple combinatorics question.

Okay bro, we understand. ;)

>>5073693
Is it simple? I can't tell because it's too ugly to bother reading. Which means that it definitely is not clear.

>>5073710
Okay. Your intellect is a great loss to all of us. ;)