The binomial coefficient — symmetry

What just happened?

We highlighted two different positions in the row — k and n−k — placed symmetrically about the centre. Their bars have identical height, because $C(n,k)=C(n,n-k)$. The whole row of Pascal’s triangle is mirror-symmetric.

The binomial coefficient:

— the number of k-element subsets of an n-element set (no repetition, order irrelevant). $n$ — the size of the set, $k$ — how many elements we choose.

Symmetry — the heart of the model:

Choosing $k$ elements “to take” is the same as pointing out the remaining $n-k$ elements “to leave out”. Each k-element choice corresponds to exactly one (n−k)-element choice — which is why there are equally many of both. It also shows in the formula: swapping $k \leftrightarrow n-k$ does not change the denominator $k!\,(n-k)!$.

Connection with Pascal’s triangle and the binomial

Row $n$ of Pascal’s triangle is, in order, $C(n,0), C(n,1), \dots, C(n,n)$ — and symmetry means it reads the same from left and from right. The same numbers are the coefficients of Newton’s binomial:

e.g. for n = 4: 1, 4, 6, 4, 1 — a palindrome. The extreme terms are always $C(n,0)=C(n,n)=1$, and the largest lies in the middle of the row.

Why you need this for the exam:

Symmetry shortens calculations: instead of computing $C(20,17)$ compute $C(20,3)$ — the same value, less multiplication. It is also a quick correctness check: if the row is not symmetric, there is an error somewhere.

Remember: $\binom{n}{k}=\binom{n}{n-k}$ — choosing k “to take” is choosing n−k “to leave out”. The Pascal row is symmetric.