Combinatorial proof for $ \frac{1}{2} \binom{2n}{n} = \binom{2n-1}{n}$

Question

I have to find a combinatorial proof for the following identity:

$$ \frac{1}{2} \binom{2n}{n} = \binom{2n-1}{n}$$

I am really not sure how to formulate the proof because I have a scalar in front of binomial, in this case, a fraction.

Answer 1

There is a identity which says that $ \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ now just compare left and right hand side combinatorial proofs will be like this first we have to choose n persons from 2n persons which is same thing as if one person is fixed then total ways are apparant from here I suppose you can complete

Answer 2

First transform the identity to $$\binom{2n}{n} = 2 \binom{2n - 1}{n}\ .$$ The left hand side is the number of ways of choosing an $n$ elements subset out of a $2n$ element set $S$. Now specify one element $r\in S$, it can be either included in the chosen subset or not. If it is not included, then we need to choose $n$ elements out of the remaining $2n-1$ elements to form the subset, of which there are $\binom{2n-1}{n}$ ways. If it is included, then we need to choose $n-1$ elements out of $S\setminus \{r\}$, or equivalently, we can choose $n$ elements out of the remaining $2n-1$ elements to exclude, of which there are also $\binom{2n-1}{n}$ ways.

Answer 3

Divide a class of $2n$ students into two teams of $n$ members each. The oldest student is present in only one of them. Of all such team formations, $\binom{2n}{n}$ in total, he is present in only half of them and absent in other half.

All possible $n$-strong teams that include the oldest student can also be generated in $\binom{2n-1}{n}$ ways by selecting $n$ of the other $2n{-}1$ students to exclude from joining a team with him.