Solution

a) For the first student, there are 5 possibilities, for the second 4 possibilities, for the third 3, for the fourth 2, and for the last one, only 1 possibility. Therefore, the total number of possibilities is $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$.

b) There is only one sequence in which they will go to the board in alphabetical order. Therefore, the number of sequences in which they will NOT go in alphabetical order is $120-1=$ 119.

c) Let's consider Arnaldo and Bernaldo as if they were just one person: ABnaldo. Thus, there would be $4 \cdot 3 \cdot 2 \cdot 1=24$ possibilities, and for each of these, we can switch the positions of Arnaldo and Bernaldo while keeping them side by side. Therefore, the total number of possibilities is $2 \cdot 24=48$.
Professor Piraldo will conduct an evaluation for 5 students in a class: Arnaldo, Bernaldo, Cernaldo, Dernaldo, and Ernaldo. This test involves calling each one to the board, once, to solve a problem each.

a) In how many different ways can the teacher call them to the board?

b) In how many of these sequences are they NOT in alphabetical order?

c) In how many of these sequences are Arnaldo and Bernaldo called in consecutive positions?