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?
 A: First bullet:  You have over-counting only if the two 5-member 'teams' are
indistinguishable. If they are committees for conducting initial phases of a job search (say: Screening, Travel arrangements, and Interviewing), then you can
use a 'multinomial coefficient':
$${12 \choose 2, 5, 5} = \frac{12!}{2! \cdot 5! \cdot 5!} = {12 \choose 2}{10 \choose 5}{5\choose 5} = 15,632.$$
choose(12,2)*choose(10,5)
## 16632
factorial(12)/(2*factorial(5)^2)
## 16632

Second bullet: If you are finding a probability, it is important
to count elements from the same sample space of equally likely elements
in both numberator and denominator.
One way to do the computation is to find that there are $10^3 = 1000$ ordered ways
in which to request three different classes. But there are only $10\cdot 9\cdot 8 = 720$ ways to choose them without conflict. So the probability of avoiding
conflict is $720/1000 = 0.720.$
Sometimes I find it helpful to do a simple simulation as a reality check
of a combinatorial model. In the R program below; $U$ is the number
of uniquely different time slots realized when 3 slots are chosen at
random (unwittingly perhaps with replacement) from among 10. With a million
iterations, you can expect about two-place accuracy. [The vector u==3 is a logical vector of a million TRUEs and FALSEs; its mean is the proportion of its TRUEs.]
set.seed(702);  m = 10^6;  slots = 1:10
u = replicate( m, length(unique(sample(slots, 3, repl=T))) )
mean(u == 3)
## 0.720409

If the courses are Math, Physics, Spanish, then the two orders MPS and SMP of requesting courses would both be counted (in numerator and in denominator). If it happens that Physics and Spanish are scheduled in the same slot, both outcomes would result in a conflict. If the college assigned these three courses to time slots at random then it seems there are 1000 equally likely possibilities in the sample space.
Note: I'm not sure what probability problem you have in mind for the problem at your first bullet, or a useful connection between the two problems. (Does your text state or imply a direct connection?) Maybe
someone else does--and will explain.
A: 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 
$$1-1 \cdot \frac{9}{10} \cdot \frac{8}{10}$$
