Probability problem confusion

Question

Hello i have a problem and I'm not sure why my solution is wrong.

Problem

There are 90 students.

We're gonna split them in 3 groups, 30 students each.

Find the probability that 2 given students A and B, will end up on the same group.

Solution from notes

If we place A on a group then there are 29 places out of 89 that we can place B such that they are on the same group. Therefore the probability is 29/89.

My solution

There are 9 permutations: {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}

that show in which group A and B could go. For example, the permutation (1,2) means that A will go to the first group and B will go to the second group.

Out of those only 3 permutations are favorable: (1,1), (2,2) and (3,3).

So the probability is 3/9 = 1/3.

Answer 1

In your calculation, cases are not equally likely. For example, you have $30\times 29$ situations for cases $(k,k)$ and $30\times30$ cases for $(k,l)$ where $k\neq l$. So, the probability will be $$\frac{30\times29\times3}{30\times29\times3+30\times30\times6}=\frac{29}{29+30\times2}=\frac{29}{89}$$