Twenty people are queuing. Two new lines are opened and 8 people rush there. At least two of these 8 end up in each queue. In how many ways can these queues be created?
If everyone was in one queue there would be $ \binom{20}{12}$ ways.
I need to figure out in how many ways I can distribute 8 people in two queues, restricted to at least two in each queue. Then multiply that with $ \binom{20}{12}$. Is that correct? How do I figure out the second part?
$20$ people are standing in a line. $8$ leave to form two new queues. The new queues can have $(2,6)$ or $(3,5)$ or $(4,4)$ or $(5,3)$ or $(6,2)$ people respectively.
You're correct in saying the result is $\binom{20}{8}$ times something.
For second part, there are $5$ cases of arranging $8$ people. The something is $5\cdot 8!$
Slightly shorter route, we have $5$ ways of arranging $8$ people from $20$. That's $$ ^{20}P _{8} \times 5$$