Row in Pascal's Triangle forms Palindromic Sequence

Theorem

Each of the rows of Pascal's triangle forms a palindromic sequence.

Proof

The $n$th row of Pascal's triangle consists of the finite sequence:

$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$

By the Symmetry Rule for Binomial Coefficients:

$\dbinom n m = \dbinom n {n - m}$

Hence we can write the $n$th row in reverse order:

\(\ds \)

\(\)

\(\ds \dbinom n n, \dbinom n {n - 1}, \dbinom n {n - 2}, \ldots, \dbinom n 2, \dbinom n 1, \dbinom n 0\)

\(\ds \)

\(=\)

\(\ds \dbinom n {n - n}, \dbinom n {n - \left({n - 1}\right)}, \dbinom n {n - \left({n - 2}\right)}, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n {n - 0}\)

\(\ds \)

\(=\)

\(\ds \dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n\)

and the sequences are seen to be the same.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $8$