Problem:
$50$ people go to teather. The price of the ticket is $5$\$. $25$ of these people have only a $5$\$ note in their wallet, while the other $25$ only have a $10$\$ banknote. The cashier doesn't have any cash at the beginning of the queue. I would like to know how many queues exist such that the cashier always has the possibility to give the rest.
For instance: if the first person pays with a $10$\$ note, the queue does not work, so we know that the first person who pays, needs to pay with a $5$\$ note, because if the first one comes with $5$\$, then the second one can pay with $10$\$, since in this case the cashier would have the rest.
So, how many different queues are possible?
My thoughts:
I thought i would have found the solution by finding the number of arrangements of $50$ elements by using only $\{0,1\}$, where $0$ stands for a person who pays with a $5$\$ note, while $1$ stands for a person who pays with a $10$\$ note. In this way I know that the number of $0$s must always be bigger or equal to the number of $1$s, if you read the arrangement from left to right.
I also found that this number must be smaller than $\dfrac{1}{2}\begin{pmatrix} 50 \\ 25 \end{pmatrix}$, so a half of total number of arrangements.
Can somebody help me?
