Combinatorics (combinations problem) How many ways are there to pick a group of $4$ people from $10$ people (each of a different height) and then pick a second group of $3$ other people such that all the people in the first group are taller than all the people in the second group? (Hint: Consider two  ways)
I try to illustrate it to sort the shortest to the tallest, then I found out four different cases. But my lecturers mean to use 2 ways, can anyone help me??

 A: Just pick $7$ people from $10$, and let the $3$ shortest ones be called the second group. This can be done in $\binom{10}{7}$ ways.
Remark: The cases approach of the post does some double-counting. One can adjust it, by organizing by "shortest in the group of $4$," If the shortest in that group is to be say Person $6$, then we need to choose three from the $5$ tallest. 
A: Let's number the people by $1,\dots,10$ in order of length. So $1$
is the smallest. The shortest of the first group chosen must belong
to $\left\{ 4,5,6,7\right\} $. This because at least $3$ persons are
shorter and at least $3$ persons are taller.
If $4$ is the shortest of the first group then there are $\binom{6}{3}\times\binom{3}{3}=20$
possibilities.
If $5$ is the shortest of the first group then there are $\binom{5}{3}\times\binom{4}{3}=40$
possibilities.
If $6$ is the shortest of the first group then there are $\binom{4}{3}\times\binom{5}{3}=40$
possibilities.
If $7$ is the shortest of the first group then there are $\binom{3}{3}\times\binom{6}{3}=20$
possibilities.
The first factor corresponds with the first group (where the smallest
is determined allready) and second with the second group.
In total there are $120$ possibilities. 
In your calculation there
are possibilities that are counted twice.
