Each of the bicycle plates in Quixajuba contains three letters. The first letter is chosen from the set $\mathcal{A}=\{\mathrm{G}, \mathrm{H}, \mathrm{L}, \mathrm{P}, \mathrm{R}\}$, the second letter is chosen from the set $\mathcal{B}=\{\mathrm{M}, \mathrm{I}, \mathrm{O}\}$, and the third letter is chosen from the set $\mathcal{C}=\{\mathrm{D}, \mathrm{U}, \mathrm{N}, \mathrm{T}\}$.

Due to the increase in the number of bicycles in the city, it was decided to expand the number of possible plates. It was determined to add two new letters to only one of the sets or one new letter to two of the sets.

What is the maximum number of new plates that can be made when the two new letters are added?
Initially, it is possible to register $5 \times 3 \times 4=$ 60 bicycles. Let's analyze the two possible situations:

- We increase two letters in one of the sets. With this, we can have
Suggestion: Calculate the initial number of plates that can be made with the elements of sets $\mathcal{A}, \mathcal{B}$, and $\mathcal{C}$, and then recalculate by analyzing the various possibilities of increasing the elements of the sets by 1 or 2.

| $\mathcal{A} \times \mathcal{B} \times \mathcal{C}$ | Number of Plates |
| :---: | :---: |
| $7 \times 3 \times 4$ | 84 |
| $5 \times 5 \times 4$ | 100 |
| $5 \times 3 \times 6$ | 90 |

Thus, with the modification shown, the maximum number of new plates is $100-60=40$.

- Increase one letter in two of the sets. With this, we can have

| $\mathcal{A} \times \mathcal{B} \times \mathcal{C}$ | Number of Plates |
| :---: | :---: |
| $6 \times 4 \times 4$ | 96 |
| $6 \times 3 \times 5$ | 90 |
| $5 \times 4 \times 5$ | 100 |

In this case, the maximum number of new plates is also 40.

Suggestion: In item (b), consider the players who are eliminated instead of those who advance to the next rounds.