Birthday paradox combinatorics Here's a problem in Harvard's STAT 110 Probability course:
"3. A college has 10 (non-overlapping) time slots for its courses, and blithely assigns courses to time slots randomly and independently. A student randomly chooses 3 of the courses to enroll in (for the PTP, to avoid getting fined). What is the probability that there is a conflict in the student’s schedule?"
It's similar to the Birthday Paradox. Obviously the answer is $$
P(conflict) = 1 - \frac{10}{10} \times \frac{9}{10} \times \frac{8}{10}=0.28
$$
I'm new to combinatorics and I want to get a real intuition for it, so I tried this as well: $$
P(conflict)=\frac{\binom{n+k-1}{k} - \binom{n}{k}}{\binom{n+k-1}{k}}=
0.\overline{45}
$$
Why doesn't this work? The numerator counts the number of class combinations that conflict, and the denominator counts the number of all class combinations. What am I missing?

I was using the calculations that disregard order, but order matters in this question because the classes are obviously different from one another. Thus,
$$
P(conflict)=\frac{10^3-P(10,3)}{10^3}=0.28,
$$
which is the correct answer.
