Proving an identity relating to binomial coefficients This question is from the book Discrete mathematics and its applications, by K. Rosen, 6th chapter and 27th question.

Show that if n is a positive
integers $2C(2n, n+1) + 2C(2n, n) = 
  C(2n+2,n+1)$

I know how to prove this algebraically, applying the formula and this is how it is also given in hints. Since I want to understand counting arguments I am thinking to give a proof of this using combinatorial arguments (double counting).
The right side is the number of ways to select n+1 elements from 2n+2 elements. But I am not able to get an argument for the left side. I'm not getting how can I argue that I need to choose n or n+1 elements from 2n elements twice.
Any other (counting) approaches are also fine. Thank you.

Perhaps this answer is cheating?
First use a counting argument to prove:
$$
\binom{m}{r} = \binom{m - 1}{r} + \binom{m - 1}{r - 1}
\qquad (m > r \geqslant 1).
$$
Apply this identity twice, first taking $m = 2n + 2$ and
$r = n + 1$:
$$
\binom{2n + 2}{n + 1} = \binom{2n + 1}{n + 1} + \binom{2n + 1}{n} =
2\binom{2n + 1}{n + 1},
$$
and then taking $m = 2n + 1$ and $r = n + 1$:
$$
\binom{2n + 1}{n + 1} = \binom{2n}{n + 1} + \binom{2n}{n}.
$$
We used the result that
$$
\binom{2n + 1}{n} = \binom{2n + 1}{n + 1}.
$$
This is a special case of
$$
\binom{m}{r} = \binom{m}{m - r} \qquad (m \geqslant r \geqslant 0),
$$
which can also be proved by a counting argument.
I'll get me coat. :)
