# Pascal’s triangle is symmetrical along its central column

As a consequence of Pascal’s rule, we see that Pascal’s triangle is symmetrical along its central column (the column containing the central binomial coefficients). Expressing individual values in Pascal’s triangle $T$ as $T(n,k)$, with $n$ and $k$ being integers obeying the relation $-1, this means that each $T(n,k)=T(n,n-k)$.

Since Pascal’s triangle is essentially a table in which to look up binomial coefficients,

 $T(n,k)=\binom{n}{k}.$

From Pascal’s rule it follows that $T(n,k)=T(n-1,k-1)+T(n-1,k)$.

Obviously $T(0,k)=1$ because there is only one way to choose no items from a collection of $k$ items; likewise, $T(k,k)=1$ because there is only one way to choose $k$ items from a collection of $k$ items. Therefore, the leftmost and rightmost column of Pascal’s triangle are filled with 1’s. Almost as obvious is the fact that $T(1,k)=k$ because there are $k$ ways to choose just one item from a collection of $k$ items; likewise, $T(k-1,k)=k$ because there are $k$ ways to choose all but one item from a collection of $k$ items since leaving out one item in turn can only be done $k$ times in such a collection.

From the foregoing, row 1 of Pascal’s triangle is 1, 1, row 2 is 1, 2, 1 and row 3 is 1, 3, 3, 1. From Pascal’s rule it follows that even-numbered rows (with an odd number of columns, and their highest, central value at $T(\frac{k}{2},k)$) will be symmetrical along the central value if the previous row was also symmetrical, while odd-numbered rows (with an even number of columns, and the highest, central value at both $T(\frac{k-1}{2},k)$ and $T(\frac{k+1}{2},k)$ will be symmetrical about the central values if the previous row was symmetrical. Since the first three rows are symmetrical, all the following rows are also symmetrical.

Title Pascal’s triangle is symmetrical along its central column PascalsTriangleIsSymmetricalAlongItsCentralColumn 2013-03-22 19:00:14 2013-03-22 19:00:14 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Corollary msc 05A19