35. [25] Let $P$ denote the set of all subsets of $\{1, \ldots, 23\}$. A subset $S \subseteq P$ is called good if whenever $A, B$ are sets in $S$, the set $(A \backslash B) \cup(B \backslash A)$ is also in $S$. (Here, $A \backslash B$ denotes the set of all elements in $A$ that are not in $B$, and $B \backslash A$ denotes the set of all elements in $B$ that are not in $A$.) What fraction of the good subsets of $P$ have between 2015 and 3015 elements, inclusive?
If your answer is a decimal number or a fraction (of the form $m / n$, where $m$ and $n$ are positive integers), then your score on this problem will be equal to $\max \{0,25-\lfloor 1000|A-N|\rfloor\}$, where $N$ is your answer and $A$ is the actual answer. Otherwise, your score will be zero.
Answer: $\quad \frac{18839183877670041942218307147122500601235}{4769168484048619242209570178451249271212} \approx 0.3950203047068107$ Let $n=23$, and $\ell=$ $\lfloor n / 2\rfloor=11$.
We use the well-known rephrasing of the symmetric difference $((A \backslash B) \cup(B \backslash A))$ in terms of addition modulo 2 of "indicators/characteristic vectors". So we simply want the number $\binom{n}{\ell}_{2}$ of dimension $\ell$ subspaces of the $F:=\mathbb{F}_{2}$-vector space $V:=\mathbb{F}_{2}^{n}$. Indeed, good subsets of $2^{d}$ elements simply correspond to dimension $d$ subspaces (in particular, good subsets can only have sizes equal to powers of $|F|=2$, and $2^{\ell}$ is the only power between 2015 and 3015 , inclusive).
To do this, it's easier to first count the number of (ordered) tuples of $\ell$ linearly independent elements of $V$, and divide (to get the subspace count) by the number of (ordered) tuples of $\ell$ linearly independent elements of any $\ell$-dimensional subspace of $V$ (a well-defined number independent of the choice of subspace).
In general, if we want to count tuples of $m$ linearly independent elements in an $n$-dimensional space (with $n \geq m$ ), just note that we are building on top of (0) (the zero-dimensional subspace), and once we've chosen $r \leq m$ elements (with $0 \leq r \leq m-1$ ), there are $2^{n}-2^{r}$ elements linearly independent to the previous $r$ elements (which span a subspace of dimension $r$, hence of $2^{r}$ "bad" elements). Thus the number of $m$-dimensional subspaces of an $n$-dimensional space is
$$
\binom{n}{m}_{2}:=\frac{\left(2^{n}-2^{0}\right)\left(2^{n}-2^{1}\right) \cdots\left(2^{n}-2^{m-1}\right)}{\left(2^{m}-2^{0}\right)\left(2^{m}-2^{1}\right) \cdots\left(2^{m}-2^{m-1}\right)}
$$
a "Gaussian binomial coefficient."
We want to estimate
$$
\frac{\binom{n}{\ell}_{2}}{\sum_{m=0}^{n}\binom{n}{m}_{2}}=\frac{1}{2} \frac{\binom{23}{11}_{2}}{\sum_{m=0}^{11}\binom{23}{m}_{2}}
$$

To do this, note that $\binom{n}{m}_{2}=\binom{n}{n-m}_{2}$, so we may restrict our attention to the lower half. Intuitively, the Gaussian binomial coefficients should decay exponentially (or similarly quickly) away from the center; indeed, if $m \leq n / 2$, then
$$
\binom{n}{m-1}_{2} /\binom{n}{m}_{2}=\frac{\left(2^{m}-2^{0}\right) \cdot 2^{m-1}}{\left(2^{n}-2^{m-1}\right)} \approx 2^{2 m-1-n} .
$$

So in fact, the decay is super-exponential, starting (for $n=23$ odd and $m \leq \ell=11$ ) at an $\approx \frac{1}{4}$ rate. So most of the terms (past the first 2 to 4 , say) are negligible in our estimation. If we use the first two terms, we get an approximation of $\frac{1}{2} \cdot \frac{1}{1+\frac{1}{4}}=\frac{2}{5}=0.4$, which is enough for 20 points. (Including the next term gives an approximation of $\frac{32}{81} \approx 0.3950617$, which is good enough to get full credit.)
To compute the exact answer, we used the following python3 code:
```
from functools import lru_cache
from fractions import Fraction
@lru_cache(maxsize=None)
def gauss_binom(n, k, e):
    if k  n:
        return 0
    if }\textrm{k}==0\mathrm{ or }\textrm{k}==\textrm{n}\mathrm{ :
        return 1
    return e ** k * gauss_binom(n - 1, k, e) + \
            gauss_binom(n - 1, k - 1, e)
N}=2
K = 11
good = gauss_binom(N, K, 2)
total = sum(gauss_binom(N, i, 2) for i in range(N + 1))
print(Fraction(good, total))
print(float(good/total))
```