<s>[INST] Splitting a population, probability of 2 people landing in same subpopulation. 
$10$ people has been split into $3$ groups $A,B,C$ of $5,3,2$ people respectively. What is the probability that $2$ predetermined people $x,y$ land in the same group?

My attempt:
There is $\binom{10}{5, 3, 2} = \frac{10!}{5!3!2!}$ ways to split the population into mentioned above subpopulations. 
Let's count all the partitions in which $x,y\in A$. 
There's $\binom{10-2}{3}$ ways to choose the rest of people in $A$, $\binom{5}{3}$ ways to choose people in $B$ from the remaining 5, and $\binom{2}{2}$ ways to choose people in $C$. Repeating the above for the cases $x,y\in B$ and $x,y \in C$ we get:
$$P(A)=\frac{  \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5}}{\binom{10}{5,3,2}},$$
which is wrong.
The given answer is 
$$P(A)=\frac{  \binom{8}{3}\binom{2}{2} +\binom{8}{1}\binom{2}{2}+\binom{2}{2}}{\binom{10}{5,3,2}},$$
and I'm having a hard time engineering it back. Could anyone point where I made mistake?
 [/INST] I also think that you are correct. I get the same answer by just partitioning the remaining eight people ...
$$N(A)= \binom8{3,3,2}+\binom8{5,1,2}+\binom8{5,3,0}
\\= \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5}
\\=784$$
</s>