Sampling with and without replacement; Overcounting I am really having a lot of trouble with counting questions in my intro Probability course. An issue I am having now:
• How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?
For this question, I had no idea how to start. The answer said to pick 2 people for the 2-man team, then 5 for the 5-man team, hence ${12 \choose 2} \cdot {10\choose 5}.$ But this overcounts by a factor of 2 so we have to divide by 2. I understood this overcounting as there being two ways to arrange the 5-man teams.
• 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. What is the probability that there is a conflict in the student’s schedule?
I next tried this question. I first started by counting the max number of combinations of 3 courses, hence $10^3.$ I then thought that there would similarly be overcounting (since it doesn't matter the order of the courses) and then divided by 3!. From the answer key, this is wrong and I have no idea why.
Can someone please help me here? And do you guys have any strategies when approaching counting questions that could help?

For the second problem, it may be easier to find the probability that there is no conflict in the student's schedule, and then subtract from 1.
For the first class there is no possible conflict, so the probability of no conflict is 1.  For the second class, there is one bad day, so the probability of no conflict is $9/10$.  For the third class, if there has been no conflict so far, there are two bad days, so the probability of no conflict is $8/10$. So the probability of no conflict in all three classes is
$$1 \cdot \frac{9}{10} \cdot \frac{8}{10}$$
and the probability that there is a conflict is 
$$1-1 \cdot \frac{9}{10} \cdot \frac{8}{10}$$
