Method 1: We begin by arranging three blue balls and three red balls in a row. There are $\binom{6}{3}$ such arrangements since we must choose which three of the six spaces will be occupied by the blue balls. This creates seven spaces in which we can place three green balls, five spaces between successive balls and two at the ends of the row. For instance,
$$\square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square$$
We now wish to insert three green balls so that no two of them are adjacent. To do so, we must choose three of these seven spaces, which can be done in $\binom{7}{3}$ ways. One such arrangement is
balls and two at the ends of the row. For instance,
$$\color{green}{\bullet} \color{red}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{red}{\bullet} \color{blue}{\bullet} \color{red}{\bullet}$$
The green balls represent the positions that can be occupied by Alvin, John, and Albert; the blue balls represent the positions of the other three people; the red balls represent the positions of the three unoccupied seats. We can arrange Alvin, John, and Albert in the places occupied by green balls in $3!$ ways. We can arrange the remaining three people in the positions occupied by the blue balls in $3!$ ways.
Consequently, the number of permissible seating arrangements is
$$\binom{6}{3}\binom{7}{3}3!3!$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $\binom{9}{3}$ ways of choosing the locations of the empty seats and $6!$ ways of arranging the six people in the remaining seats. From these seating arrangements, we must exclude those arrangements in which the three students who do not wish to sit in adjacent seats sit in adjacent seats.
If two of Alvin, John, and Albert sit together, we have eight objects to arrange, the pair of seats in which we will place those students, the three empty seats, and the other four students. There are $\binom{8}{3}$ ways to choose three of the eight positions for the empty seats, $\binom{3}{2}$ ways to choose two of the three students to sit together, $5!$ ways to arrange the remaining objects, and $2!$ ways to arrange the pair of chosen students in the designated pair of seats.
$$\binom{8}{3}\binom{3}{2}5!2!$$
Subtracting this from the total removes those cases in which all three students sit together twice, once when we designate the leftmost two students as the pair and once when we designate the rightmost two students as the pair. Since we only wish to exclude these arrangements once, we must add them back.
If all three students sit together, we have seven objects to arrange, the three empty seats, the trio of seats occupied by Alvin, John, and Albert, and the other three people. We can select three of these seven positions for the empty seats in $\binom{7}{3}$ ways. We can arrange the remaining four objects in $4!$ ways. We can arrange Alvin, John, and Albert within the designated trio of seats in $3!$ ways. Hence, the number of seating arrangements in which Alvin, John, and Albert sit together is
$$\binom{7}{3}4!3!$$
By the Inclusion-Exclusion Principle, the number of permissible seating arrangements is
$$\binom{9}{3}6! - \binom{8}{3}\binom{3}{2}5!2! + \binom{7}{3}4!3!$$
Best Answer
Your mistake is in thinking that you get to pick which $5$ students always sit in the front row: you don’t. You have no choice in the matter: it’s a definite set of $5$ students who always sit in the front row. Similarly, it’s a definite set of $4$ students who always sit in the back row. Then there are $5$ students who sometimes sit in the front row and sometimes in the back.
And yes, this is implied by the wording of the problem.