$6$ persons stand at random in a queue for buying cinema tickets individually 
$6$ people stand at random in a queue for buying cinema tickets individually. Three of them have only a five rupees note each while each of the other $3$ has a ten rupees note only. The booking clerk has an empty cash box. Find the probability that the six people will get tickets, each paying  five rupees.

Attempt: Let $S$ be the sample space and $E$ be the required events, then the total number of arrangements of $6$ people equals $6! = 720$. Now for favorable number of cases, let $a$ denote a person having $5$ rupees note and $b$ denote the person having $10$ rupees:
$\bullet\; $ Arrangements as $a \; (ba)\; (ba)\; b$ 
$\bullet\; $ Arrangements as $a\; (ab)\; (ab)\; b$
How can I calculate the number of favorable cases?
 A: If $F$ denotes a person with five rupee note and T denotes the person with a ten rupee note, then we need arrangements such that the number of F is more than the number of T at any point in the queue. This is equivalent to finding the number of paths from the lower left corner to the upper right corner in the figure that lies below the diagonal line. Counting the number of ways or reaching each grid point, the number of favorable cases is 5.
Arrangements of 6 persons in the queue is $\frac{6!}{3!3!}$, since we are interested only in all getting tickets. Thus the probability is $\frac{5}{20} = \frac{1}{4}$.
More generally, if there are $n$ persons with five rupee notes and $n$ persons with ten rupee notes, then the number of favorable cases is $\frac{1}{n+1}\binom{2n}{n}$, the $n$ th Catalan number. 
A: There are not too many cases to just count. You must always start with a $5$ and end with a $10$, and make sure that there are at least as many with a $5$ as with a $10$ at any point in time. Valid cases are thus:
$$5, 5, 5, 10, 10, 10$$
$$5, 5, 10, 5, 10, 10$$
$$5, 5, 10, 10, 5, 10$$
$$5, 10, 5, 5, 10, 10$$
$$5, 10, 5, 10, 5, 10$$
Since there are $3! \cdot 3! = 6 \cdot 6 = 36$ permutations for each valid case, the number of favorable cases is $5 \cdot 36 = 180$, and the probability of having a valid row of people equals:
$$\frac{180}{720} = \frac{1}{4}$$
