Number of parenthesizations

Task number: 3453

How many distinct arrangements are there of \(n\) pairs of parentheses such that the parentheses can be correctly paired (i.e. good parenthesizations)?

  • Hint

    Try to describe the good (and bad) parenthesizations as paths in an \(n\times n\) grid and then look for a property of the bad paths which will let you count them.

  • Solution

    We can represent a parenthesization of \(n\) pairs of parentheses as a path of length \(2n\) in an \(n\times n\) grid (i.e. with \(n^2\) squares) from the upper left corner to the lower right corner, where an opening parenthesis '(' indicates a step to the right and a closing parenthesis ')' indicates a step downward. The number of such paths, i.e. of such parenthesizations is \(\binom{2n}{n}\)

    The bad parenthesizations are the ones which fall below the diagonal – at that moment the number of closing parenthses ')' used exceeds the number of opening parentheses '('.

    We determine the number of bad paths as follows. We divide a bad path \(P\) into two segments, namely segment \(A\) from the start to the moment when the path falls below the diagonal, and the remaining segment \(B\). We take the mirror image of segment \(A\), yielding a segment \(A'\), whose extension via \(B\) yields a path \(P'\).

    This path \(P'\) is in an \((n-1)\times (n+1)\) grid. The relationship \(P \leftrightarrow P'\) is a bijection between bad paths in the original grid and all paths in the new grid, because we can also uniquely divide the path \(P'\) into segments \(A'\) a \(B\). Segment \(A'\) ends as soon as the path crosses to the right of a diagonal line leading from the upper left corner of the new grid. Such a moment must always occur, because the new grid is wider than it is long.

  • Answer

    The number of possible parenthesizations is \(\binom{2n}{n}-\binom{2n}{n-1}=\frac{1}{n+1}\binom{2n}{n}\).

Difficulty level: Hard task
Reasoning task
Solution require uncommon idea
Cs translation
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