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 , 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,

$$T(n,k)=\left(\genfrac{}{}{0pt}{}{n}{k}\right).$$

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.