Can't understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct I'm following my teacher's notes and I cannot understand how $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}(2k+4)!(k+2)!^2$ is correct. Shouldn't it be $\frac{1}{2}{ 2k+4 \choose k+2} = \frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}$ ?
Not sure if relevant, but the exercise (with which I don't need help with!) is to prove by induction ${0 \choose 0} + {2 \choose 1} + ... + {2n \choose n} < \frac{1}{2} {2n + 2 \choose n + 1}$ where $n \geqslant 2$.
Thanks

Indeed by definition of the binomial coefficient
$$\frac{1}{2}\binom{2k+4}{k+2}=\frac{1}{2}\frac{(2k+4)!}{(k+2)!^2}.$$
Plugging in any value, for example simply $k=0$, shows that
$$\frac{1}{2}\frac{(2k+4)!}{(k+2)!(k+2)!}\neq\frac{1}{2}(2k+4)!(k+2)!^2,$$
so this is indeed a misprint as suggested in the comments.
