Probability of three friends being in the same group -- Confusion on Counting Problem: 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?
So I started by noting that there are $10 \choose 5, 5$ = $10 \choose 5$  ways of picking groups of 5 and that's my denominator. My numerator is $7 \choose 2$ because I want to pick two other individuals out of the remaining seven if you were to fix me and my two friends in one group together. The answer, however has $2$$7 \choose 2$ as my numerator. Why is there a $2$?
Similarly, I saw in Probability of two friends being in the same group (same problem but two friends instead of three) that one of the answers counts $8 \choose 3$ and $8 \choose 5$ because it counts for picking three people to be in your group as well as picking 5 people to have neither friend. Why are these two cases different? Isn't $8 \choose 3$ accounting for both cases?
 A: While you have got the answer to your query re $2$ in the numerator, I'd like to recommend a much simpler way to solve the problem:
You can be in any group, and now there are 4 slots left in your group in a total of $9$ vacant
Thus $P$(your $2$ friends are in your group) = $\dfrac 49\cdot\dfrac38 = \dfrac16$
A: The factor of $2$ in the numerator comes from a factor of $1/2$ in the denominator.  
If the groups were labeled, then there would be $\binom{10}{5}$ ways of selecting the two groups.  However, if the two groups are unlabeled, we have counted each group twice, once when we select a particular group of five people and once when we select its complement.  Hence, the number of ways to select two unlabeled groups of five people from a group containing $10$ people is 
$$\frac{1}{2}\binom{10}{5}$$
Another way to see this is to suppose that Amanda is one of the ten people.  A particular selection of two groups of five people is determined by selecting which four of the other nine people are in Amanda's group.  Therefore, there are 
$$\binom{9}{4}$$
ways of selecting two groups of five people from a group of ten people.  As you can verify by direct calculation or by using Pascal's Identity, $$\binom{9}{4} = \binom{9}{5} = \frac{1}{2}\binom{10}{5}$$ 
You are correct that the numerator is $\binom{7}{2}$ since the group that contains the three friends is determined by which two of the remaining seven people are in the group.  Thus, the probability that the three friends are in the same group is 
$$\frac{\binom{7}{2}}{\binom{9}{4}} = \frac{\binom{7}{2}}{\frac{1}{2}\binom{10}{5}} = \frac{2\binom{7}{2}}{\binom{10}{5}}$$
