We need the formulas for combinations and permutations
[tex]\begin{gathered} \text{nCk}=\frac{n!}{k!(n-k)!} \\ \text{nPk}=\frac{n!}{(n-k)!} \end{gathered}[/tex]Replacing the information we have
[tex]\begin{gathered} 9C3=\frac{9!}{3!(9-3)!}=\frac{9!}{3!6!}=84 \\ 12P3=\frac{12!}{(12-3)!}=\frac{12!}{9!}=1320 \end{gathered}[/tex]Then 9C3=84 and 12P3=1320.
