In this case, we have to use permutations to solve this problem, we have 3 options for 3 positions, use permutation this way:
[tex]\begin{gathered} 3P3=\frac{3!}{(3-3)!} \\ 3P3=\frac{3!}{0!}=6 \end{gathered}[/tex]The coins can be given in 6 different ways.