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${n! \over k!(n-k)!} = {n \choose k}$

\begin{equation} x=\sqrt{b} \end{equation}

If $a^2=b$ and $ b=2 $, then the solution must be either $$ a=+\sqrt{2} $$ or \[ a=-\sqrt{2} \].

msg = "Hello World"

msg

Hello World