Gauging level of math required and trying to learn more about this probability / combinatorics / statistics problem. A manager must form a team of 5 from among 5 employees from group A and 6 employees from group B. If all employees have an equal chance of being selected, what is the approximate probability that a randomly selected group of 5 will consist of 2 employees from group A and 3 employees from group B? Note that a friend has already provided me with a fairly complex combinatorics solution to this problem but he was not able to simplify on how the solution was obtained. I would definitely like learn more on how to solve this and I'm very curious to see if there are some more simpler ways to solve this problem. Any other ideas or suggestions are also welcomed.

There are ${5+6 \choose 5} = 462$ equally likely ways of choosing the team of five from the eleven people

Of these, there are ${5 \choose 2}{6 \choose 3}=10 \times 20 = 200$ ways of choosing two from the As and three from the Bs

So the probability is $\frac{200}{462} \approx 0.4329$

Is this the "fairly complex combinatorics solution"?