Answer
255
Work Step by Step
$n!=n(n-1)(n-2)\cdot...\cdot 2\cdot 1$
$0!=1$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
-------------------------
$\left(\begin{array}{l}
10\\
2
\end{array}\right) + \displaystyle \left(\begin{array}{l}
10\\
6
\end{array}\right)=\frac{10!}{2!8!}+\frac{10!}{6!4!}$
$=\displaystyle \frac{10\cdot 9}{2}+\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2}$
$=45+210=255$
