Linear arrangement - people sitting in a row

Question

There are $20$ people from $4$ cities - $8$ from A, $4$ from B, $4$ from C, $4$ from D. They are randomly being ordered in a row. The question is how many orders exist where all residents of B are standing in the left half of the row?

My thought process is that first we order the $16$ other people for which there are $16!$ options. Then we need to place the $4$ other people into the $7$ spaces on the left side of the row.

I'm not sure if the answer should be $16!\binom{10}{6}$, or we should consider the order in which the B people are placed in the row so that the answer will be $16!4!\binom{10}{6}$.

Any advice will be appreciated.

Answer 1

You have to consider the order in which the people from city B are placed in the row.

Choose four of the first ten positions for the people from city B. Arrange them in those four positions. Arrange the remaining sixteen people in the remaining sixteen positions. Doing so yields $$\binom{10}{4}4!16!$$ in agreement with your second answer. Evidently, you chose which six of the first ten positions would be occupied by the people from other cities, which is equivalent to choosing which four of those positions will be occupied by residents of city B.

Answer 2

This is correct assuming that each individual person is considered distinct (as is common for problems like these).