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?
Best Answer
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) = $\dfrac 49\cdot\dfrac38 = \dfrac16$