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??


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. 
