Use combinations to explain why $C(n, k) = C(n, n-k)$ for any $n, k ∈ ℕ, 0 ≤ k ≤ n$.

Question

• Previous subquestion:

Use the formula $\begin{pmatrix}n\\r\end{pmatrix}$ = $\frac{n!}{r!(n-r)!}$ to show that $\begin{pmatrix}n\\k\end{pmatrix}$ = $\begin{pmatrix}n\\n-k\end{pmatrix}$ for any n, k ∈ ℕ, 0 ≤ k ≤ n.

• My answer:

$\begin{pmatrix}n\\k\end{pmatrix}$ = $\frac{n!}{k!(n-k)!}$

$\begin{pmatrix}n\\n-k\end{pmatrix}$ = $\frac{n!}{(n-k)!(n-(n-k))!}$

$\begin{pmatrix}n\\n-k\end{pmatrix}$ = $\frac{n!}{(n-k)!(n-n+k)!}$

$\begin{pmatrix}n\\n-k\end{pmatrix}$ = $\frac{n!}{(n-k)!k!}$

∴ $\begin{pmatrix}n\\k\end{pmatrix}$ = $\begin{pmatrix}n\\n-k\end{pmatrix}$

• The question I'm having trouble with:

Use combinations to explain why this is true.

Answer 1

Every choice of $k$ objects necessarily determines the complementary choice of $n-k$ objects.

If there are ten bottles on a wall, when you take three down, seven are left. The number of ways of taking three is the same as the number of ways of leaving seven. When you take seven down, there are three left. Whether you consider this as $\binom{10}{3}$ or $\binom{10}{7}$ you get the same answer.

Answer 2

$\binom nk$ is by definition the number of ways of choosing $k$ elements out of $n$. There are two ways of doing so :

• Chose $k$ elements, keep them and put the others away.

• Chose $n-k$ elements, put them away and keep the remaining $k$ ones.

Therefore there are as many ways of choosing $k$ elements among $n$, and choosing $n-k$ elements among $n$.