# Cinema theatre question [duplicate]

There are 10 people waiting at the cinema theatre. 5 have Rs 50, 5 have Rs 100. The cost of the ticket is Rs 50. In how many ways can they stand in the queue such that there is no problem of change??

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It’s possible to solve this problem by brute force, but the number is large enough that you’ll really want a better approach. Your problem is the case $n=5$ of the general problem in this question, and the answer that I gave there is essentially the same as the one that I’d give here. –  Brian M. Scott Feb 27 '13 at 12:07

## marked as duplicate by Andreas Caranti, Asaf Karagila, Chris Eagle, rschwieb, Davide GiraudoFeb 27 '13 at 14:45

This is the same as the number of ways of balancing parentesis in a string of 10: Each '(' corresponds to somebody paying with Rs.50, an ')' to somebody paying with Rs.100 (you require that at each point there are no more ')' than '(' have shown up). This leads to Catalan numbers, your answer is $C_5 = \frac{1}{6} \binom{10}{5} = 42$.