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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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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 $C_5 = \frac{1}{6} \binom{10}{5} = 42$.

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I had a tough time to get the bijective proof of the above question, please explain the solution using bijection. –  user64189 Feb 28 '13 at 7:56
To me the biyection balanced parenteses <--> order in the queue is clear from the above. What exactly has you confused? –  vonbrand Feb 28 '13 at 9:59
the answer is C5 but not C10.@vonbrand –  user64189 Feb 28 '13 at 15:14
i indirectly asked for a proof of the above formula. –  user64189 Feb 28 '13 at 15:22
@user64189, you are absolutely right. Fixing. Thanks! –  vonbrand Feb 28 '13 at 16:12

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