Let the number of subsets of the set (of $2n+1$ elements) which contain at most $n$ elements be $\mathrm{M}$.
$$\therefore \mathrm{M}=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n}$$
Now we know from binomial theorem that
$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\cdots+\binom{n}{n}x^n \tag{i}$
Besides $\binom{n}{r}=\binom{n}{n-r}$
From no$\mathrm{(i)}$, putting $x=1$ and $n \to (2n+1)$ yields
$\scriptsize \begin{align}(1+1)^{2n+1}&=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n-1}+\binom{2n+1}{n}+\binom{2n+1}{n+1}+\binom{2n+1}{n+2}+\cdots+\binom{2n+1}{2n+1}\\ \Rightarrow 2^{2n+1}&=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n-1}+\binom{2n+1}{n}+\binom{2n+1}{n}+\binom{2n+1}{n-1}+\cdots+\binom{2n+1}{0}\\ &=\mathrm{M}+\mathrm{M}=2\mathrm{M}\\ \therefore \mathrm{M} &=\frac{2^{2n+1}}{2}=2^{2n}\end{align}$
So the correct answer is D.