When choosing $3$ different elements out of $8$ you have $\frac{(8!)}{(3!)}$ options.

This is called the number of 3-permutations of 8.

When choosing $3$ elements out of $8$ and their mutual order is not relevant there are  $\left( \begin{array}{c} 8\\ 3\end{array}\right)=\frac{ 8\cdot 7\cdot \; 6}{3 !}$ options.

This is called the number of 3-combinations of 8.