A set contains $2n+1$ elements. Find the subsets of the set containing atmost $n$ elements.
Since a subset may be formed by taking $0,1,2,....n$ elements , we may form a subset in the following ways $\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}........+\binom{2n+1}{n} = 2^{2n+1}$ ways. But actually in answer is $2^{2n}$. Why do we divide $2^{2n+1}$ by $2$ ?
You want to calculate $$\mathrm S=\binom{2n+1}{0}+ \binom{2n+1}{1}+\binom{2n+1}{2} +\ldots+ \binom{2n+1}{n} \tag1$$
We've
$$\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\ldots +\binom{2n+1}{2n+1}=2^{2n+1}$$
Now let's use the property $$ \binom{n}{r}=\binom{n}{n-r}$$
To transform $\mathrm S$ into $$\mathrm S= \binom{2n+1}{\color{red}{2n+1}}+\binom{2n+1}{\color{red}{2n}}+\binom{2n+1}{\color{red}{2n-1}}+\ldots + \binom{2n+1}{\color{red}{n+1}} \tag2$$
Adding $(1)$ and $(2)$, we get
$$2\mathrm S=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\ldots +\binom{2n+1}{2n+1}=2^{2n+1}$$
Thus we get $$2\mathrm S=2^{2n+1}\implies \mathrm S=2^{2n}$$
Of course. Let $A$ be a set (family) of all subset in a set $N=\{1,2,...,2n+1\}$ with at most $n$ elements and $B$ a set (family) of subsets with at least $n+1$ in N Then a map $f: A\to B$ defined with $f(X) = N\setminus X$ is bijective so $| A |=|B|={2^{2n+1}\over2}$
The equation you wrote is not true. What is true is $$\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{2n+1} = 2^{2n+1}.$$
A symmetry argument will justify why dividing by $2$ appears.
