TAOCP 1.2.6 Exercise 8

Section 1.2.6: Binomial Coefficients

Exercise 8. [00] What property of Pascal's triangle is reflected in the symmetry condition, Eq. (6)?

Verified: yes
Solve time: 35s

The symmetry condition, Eq. (6), states that

$$ \binom{n}{k} = \binom{n}{n-k}, \qquad \text{integer } n \ge 0, \ \text{integer } k. $$

This property is reflected in Pascal's triangle by the fact that each row is symmetric about its middle. Specifically, for row $n$, the entries from the left up to the center are identical to the entries from the right counting backward: the $k$th entry from the left equals the $k$th entry from the right, which corresponds exactly to the relation $\binom{n}{k} = \binom{n}{n-k}$. This symmetry arises because choosing $k$ objects from $n$ is equivalent to leaving out $n-k$ objects, so the combinatorial counts are identical.

This completes the proof.

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