Solving a problem using the Pigeonhole principle

How do I use the Pigeonhole principle to show that in a class of 25 students where every student is either a sophomore, freshman or a junior there are at least 9 sophomores or 9 seniors or 9 juniors ?

My solution: Assume by contradiction that there are 8 sophomores, 8 freshman and 8 juniors then $$ 8+8+8 = 24 \neq 25 $$ thus by contradiction there are at least 9 sophomores or 9 seniors or 9 juniors

Is this solution a correct application of the Pigeonhole principle ?

Thank you in advance guys!

This is pretty much correct.

The one nitpick I have is that, if there are not $9$ of a given year, there are at most $8$ of the given year. That is, $8$ is simply the worst case scenario.

Let a, b, c be the three values which can only be whole numbers. Then their mean is 25/3 = 8.33333 Since the mean has to lie in between the values (if not all three values are equal and not outside the range OR if all the values are identical the mean is equal to each of them (which cannot be the case here since the mean is not a positive integer), atleast one of the values must be a positive greater than 8.333... or atleast 9.