Observe the equation:

$$
\begin{aligned}
(1+2+3+4)^{2} & =(1+2+3+4)(1+2+3+4) \\
& =1 \cdot 1+1 \cdot 2+1 \cdot 3+1 \cdot 4+2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+ \\
& +3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+4 \cdot 1+4 \cdot 2+4 \cdot 3+4 \cdot 4
\end{aligned}
$$

Note that $4 \times 4=16$ products are formed when we calculate $(1+2+3+4)^{2}$ using the distributive property.

a) How many products will be formed when we calculate $(1+2+3+4)^{3}$ also using the distributive property?

b) What is the number of two-digit numbers that use only the digits 1, 2, 3, and 4?
c) What is the sum of the products of the digits of all four-digit numbers formed only by the digits $1, 2, 3,$ and 4?