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Problem:

In a queue for £1 cinema tickets, there are m people with a £1 coin and n people with a £2 coin. There is no extra change at the cinema. To buy a ticket, those with £2 coin will have to get their change. How many ways are there of queuing so that everyone gets ticket?

To make it possible, m≥n has to be satisfied.

So there are two methods to solve this in my mind. To make it easier to present, I will put apart the permutation things temporally.

One is to let possible overall number subtract the number of situation that won't meet the conditions, which is $C_{m+n}^{m} - C_{m+n}^{m+1}$.

The other looks more direct, which first select n from m people with a £1 coin to make pair with those with a £2 coin, and n vs n make it a Cn(nth Catalan Number). To sum up, the results is $C_{m}^{n}*Cn=C_{m}^{n}*C_{2n}^{n}/(n+1)$.

But unfortunately, that 2 results is not equal.

My question is: which one is right, why and why not?

Truthuleave
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2 Answers2

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Taking people to be indistinguishable,the first one is correct, it is a known method using Catalan numbers/Dyck words, as here

The second one has a fundamental error at start inasmuch as pairing the two denominations (with a £1 in the leading pair, of course) will certainly avoid shortage, but pairing is not the only way to avoid shortage, e.g. all the £1 holders could be at the head of the queue !

I think the simplest solution is using the relaxed Bertrand'sBallot Theorem where $m\geq n$ is allowed, and the probability of not coming to a stop is $\Large\frac{m+1-n}{m+1}$

so the full solution is $\Large\frac{m+1-n}{m+1}\binom{m+n}{m}$

true blue anil
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  • Thanks for your reply. Now I could get the first method. But why "pairing the two denominations (with a £1 in the leading pair, of course) is sufficient but not necessary to avoid shortage"? Could you explain that further? – Truthuleave Oct 29 '24 at 14:31
  • @VictorLi: What I meant is that there are many more possibilities, eg all £1 holders could be at the head of the queue as an extreme case, so they don't necessarily have to be in pairs. I will amend the wording in the answer, too, for greater clarity. – true blue anil Oct 29 '24 at 15:07
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You should specify in the OP that the people are indistinguishable apart from the coins they hold.

The second one cannot be correct.

Let $m=3,n=2$.

$C_{m}^{m+n} - C_{m+1}^{m+n}=5$

$C_{n}^{m}*Cn=6$

and all possible cases are

12121,12112,11221,11212,11122.

In the second method: For those chosen $n$ people with £$1$, if they happened to appear next to a not chosen person with £$1$ in a specific queue, it makes the same queue as if the other person was chosen instead.

I did not find mistakes in the first method. It follows the method as in calculating Catalan numbers.

Ma Ye
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  • Sure, the people are indistinguishable, so we could put aside permutations things. About the second method, I mean the chosen n person with £1 to be those eventually contribute change for those with £2. – Truthuleave Oct 29 '24 at 14:49