问题
I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it.
Thanks of @LutzL's suggestion. I thought of stirling's approximation before.
回答1:
The O-part should be easy. Choosing exactly n/2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not.
The Ω-part is harder. In fact, plotting 4n / binomial(2 n, n) for moderately large n I see no indication that this would flatten to stay below some constant. Speaking intuitively, the larger n is, the more special is the case when you take a random pick from n elements and coincidentially happen to choose exactly n/2 of them. That probability should tend to zero as n increases, meaning that 2n should always grow faster than n choose n/2. Are you certain you understood this part of your task correctly?
回答2:
You could use Stirlings formula for the factorials.
n! = sqrt(2*pi*(n+theta)) * (n/e)^n
where theta is between 0 and 1, with a strong tendency towards 0.
回答3:
It's not -- it's Theta(2^n/sqrt(n)), and in fact choose(n, n/2) ~ 2^n/sqrt(pi * (n/2)) as n->infinity). See https://en.wikipedia.org/wiki/Central_binomial_coefficient
来源：https://stackoverflow.com/questions/39558086/how-to-prove-binomial-coefficient-is-asymptotic-big-theta-of-two-to-the-power-n