Probability of 3 students being in the same group of 5 out of 30 students and confusing historical examples. 
Problem 1:
  There are 30 students in your class. Your science teacher chooses 5 students at random to complete a group project. Find the probability that you and your two best friends in the science class are chosen to work in the group. Explain how you found your answer.

This links to the textbook answer on page 10 of the pdf.
While I thought that the answer made sense:
P(3 friends being in the same group) = Number of groups with the 3 friends /  Number of groups of 5 possible = ${{27 \choose 2} / {30 \choose 5}}$
I came across a similar question on this forum that seems to defy the method the textbook publishes.

Problem 2: You and two of your friends are in a group of 10 people. The group is randomly split up into two groups of 5 people each. What is the probability that you and both of your friends are in the same group?

Instead of P(2 friends in same group) = Number of groups with $2$ friends/ Number of groups of $5$ possible = ${{7 \choose 2} / {10 \choose 5}}$,
There is a factor of 2 missing either by explaining that we can either consider labeled groups (counting ${7 \choose 2}$ the two are in group A and ${7 \choose 5}$ the two are in group B) or unlabeled groups (having counted the two groups twice). The answer is instead the above multiplied by $2$.
My confusion is now this inconsistency between these two situations. Why in problem 2 is ${7 \choose 2}$ in the numerator not good (or no $1/2$ in denominator) when ${27 \choose 2}$ is good in problem $1$ (or no multiplicity factor in the denominator)?
Is it possible that the textbook publisher is just wrong? Is there a special case when there are only 2 groups as opposed to the 6 in problem 1? I've considered other ways to look at each problem, but combinatorics is veritably not my strong suit, so I will leave it at that.
 A: The point is that in the first example you want the group of three to be in a particular identified group of five.
In the second example, the three people can be in either of two groups (which are not distinguished from each other). Hence the factor of $2$. There would be no factor two in this case if you wanted the three to be in the first group chosen, for example.
If, in the example with thirty people, the class was split into six groups and you wanted the three to be in one of those groups (without caring which of the six groups), the relevant factor would be six.
A: This is not really an answer but an alternative calculation (too much for a comment) that might be enlightening.
If after the choosing of $5$ students there is a random selection of $3$ students then the probability that all $3$ of them will be one of the chosen students equals:$$\frac{\binom53\binom{25}0}{\binom{30}3}=\frac{\binom53}{\binom{30}3}=\frac1{406}\approx0.002463$$
Here "you and your $2$ friends" represent the result of this selection.
Same method applied on problem 2 gives probability:$$\frac{\binom53\binom50}{\binom{10}3}+\frac{\binom50\binom53}{\binom{10}3}=2\frac{\binom53}{\binom{10}3}=\frac16\approx0.16667$$
