Video solution:

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Written solution:

Let $p$ be the number of pennies Ricardo has and let $n$ be the number of nickels he has.

We know that $$p+n=2020,$$ $$p \geq 1,$$ and $$n \geq 1$$ by the problem statement.

This means $$\begin{align*}2020 -n &=p \\ \implies 2020 -n &\geq 1.\end{align*}$$ Therefore, $1 \leq n \leq 2019.$ It follows, then, that Ricardo has $$\begin{align*}p+5n &= p+n + 4n\\ &= 2020 + 4n \end{align*}$$ cents.

Therefore, to maximize the money he has, we maximize the number of nickels he has, and minimizing the money he has involves minimizing the number of nickels he has. The maximum number of nickels he can have is $2019,$ so he can have at most $2020+2019(4)$ cents. The minimum number of nickels he can have is $1,$ so he has at least $2020+1(4)$ cents. The difference between the maximum and minimum amount of money he can have is: $$\begin{align*} &2020+2019(4) - (2020+1(4)) \\ &= 2018(4) = 8072 . \end{align*}$$

Thus, the correct answer is C.