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I need to find a recurrence relation to count the different outcomes. I'm not sure how to do this, recursively or not. This is for an undergraduate class in combinatorics. I recently learned what the Catalan Numbers are but I don't see a connection.

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    Indeed, this is one of the basic definitions of the [Catalan Numbers](https://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics). As that link points out, the reference from Richard Stanley carefully goes through all the equivalent definitions. – 2017-02-16
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    What is your favorite combinatorial representation of the Catalan numbers ? R.P. Stanley gives a list of about 200 in one of his books ! – 2017-02-16
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    @lulu ... you were 2 seconds ahead of me ... great minds & all that ... ha ha – 2017-02-16
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    @DonaldSplutterwit That is funny...almost word for word. – 2017-02-16
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    "*I recently learned what the catalan numbers are but I don't see a connection*" The very first example given [on the wikipedia page](https://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics) discusses Dyck Words: quote directly from the page "A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's" Can you replace the $X$'s and $Y$'s with heads and tails to make it match your exact wording of the problem? – 2017-02-16

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How do you know Catalan numbers? Assuming you're familiar with the definition associated with lattice paths from $(0,0)$ to $(n,n)$ that never goes above the diagonal, just think of the horizontal steps as heads and the vertical steps as tails.

If you want to find a recurrence relation, consider this: at some point from the beginning to our last coin flip, we will have at least one point where the number of heads equals the number of tails (there's always the beginning, when we have $0$ heads and $0$ tails). Consider the last time this happens (besides the end, when we have $n$ heads and $n$ tails), when we have $i$ heads and $i$ tails for some $i$ such that $0\leq i

Now what about the last $2n-2i$ flips? Note that the $2i+1$-th flip must be heads, otherwise we have $i+1$ tails and $i$ heads after the flip, and the last flip must be tails, otherwise we would have $2i$ tails and $2i-1$ heads before the flip. What can you say about the sequence of coin flips from $i+1$ heads and $i$ tails to $2i$ heads and $2i-1$ tails, then?