combinatorics - Combinatorial way of managing a store's cashier

I came across a nice question.

A certain store sells tickets for POP concerts and Rock concerts.
a POP concert ticket's price is 50 dollars, and a Rock concert ticket's price is 10 euros.
200 people are interested in buying a ticket for a POP concert, and 100 of those have a 50 dollar bill and the other 100 have a 100 dollar bill.
150 OTHER people are interested to buy a ticket for a Rock concert, where 80 of them has a bill of 10 euros and the other 70 have a bill of 20 euros.
At the start of the day, the cashier in the store is empty (no money.)
In how many ways can we arrange the 350 people in queue, that each one of them can get an exact excess to their money?

I've been thinking about it for two days now and I haven't even reached a clue on how to start solving it!

EDIT: Here's what I came up with so far:

Let us divide the problem into two sub-problems:

Product I: 50d. 200 customers. 100 have 50d (no change required) and 100 have 100d (50d change required. Let us consider the 100 who have 50d as red balls, and the 100 who have 100d as blue balls. At any given time, in a sorted line, red balls $\geq$ blue balls.

And

Product I: 10e. 150 customers. 80 have 10e (no change required) and 70 have 20e (10e change required. Let us consider the 80 who have 10e as white squares, and the 70 who have 20e as grey squares. At any given time, in a sorted line, white squares $\geq$ grey squares.

We should calculate the two above problems perhaps using Catalan's numbers (it reminds me of Catalan). And then, somehow, mix both balls and squares to be sorted in one line.

Answer

Since the store conveniently deals in euros and dollars separately, let us concentrate on arranging just the ROCK folks independently.

Since the store starts out empty, we need to arrange the people, so that at any given time the number holding $10$ euros is $\ge$ the number holding $20$ euros.

This is a classic application of Catalan numbers: you view it as a walk on grid, such that you need to stay below the diagonal.

I will leave the rest to you.