We will have the following:
a) From the information we will have that:
[tex]_{12}P_9=\frac{12!}{9!(12-9)!}\Rightarrow_{12}P_9=220[/tex]So, the total number of different boxed sets possible is 220.
b) From the information we will have that:
[tex]_{14}P_4=\frac{14!}{4!(14-4)!}\Rightarrow_{14}P_4=1001[/tex]So, the number of possible committees of size 4 is