The correct answer is $\boxed{\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \ {\text{n}} \end{array}} \right)}}{{{2^{\text{n}}}}}}$.
To see why, let’s consider the following example. Suppose we have a set of 4 elements, $A$, $B$, $C$, and $D$, and we toss a coin for each element. There are $2^4 = 16$ possible outcomes, corresponding to all the possible combinations of heads and tails. Of these outcomes, only 4 of them (HHHT, HTHH, HTH, and THHH) will result in exactly 2 heads being chosen.
In general, for a set of $2n$ elements, there are $2^{2n}$ possible outcomes. Of these, there are $\binom{2n}{n}$ outcomes that will result in exactly $n$ heads being chosen. Therefore, the probability of exactly $n$ heads being chosen is $\frac{\binom{2n}{n}}{2^{2n}}$.
We can also calculate this probability using a more formal argument. Let $X$ be the number of heads that are chosen. Then $X$ is a binomial random variable with parameters $n$ and $p = \frac{1}{2}$. The probability that $X = n$ is therefore given by
$$P(X = n) = \binom{2n}{n} \left( \frac{1}{2} \right)^n \left( \frac{1}{2} \right)^{2n-n} = \frac{\binom{2n}{n}}{2^{2n}}$$
Therefore, the probability that exactly $n$ elements are chosen is $\boxed{\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \ {\text{n}} \end{array}} \right)}}{{{2^{\text{n}}}}}}$.
The other options are incorrect. Option A is the probability that all $2n$ elements are chosen. Option B is the probability that none of the elements are chosen. Option C is the probability that at least one element is chosen.