Corollary 1

Let A be a set with n elements and let p be an integer $\leqslant n$. Then the number of subsets of A which have p elements is $\dfrac{n!}{p!(n-p)!}$.

Put $h = 2$, $p_1 = p$, $p_2 = n - p$ in Proposition 11.

¶ The number of subsets containing $p$ elements in a set of $n$ elements (where $p \leqslant n$) is denoted by $\begin{pmatrix} n \\ p \end{pmatrix}$ and is called the binomial coefficient with indices n and p. From the relation $\begin{pmatrix} n \\ p \end{pmatrix} = \dfrac{n!}{p!(n-p)!}$ it follows immediately that $\begin{pmatrix} n \\ p \end{pmatrix} = \begin{pmatrix} n \\ n-p \end{pmatrix}$.

This is also a consequence of the fact that, if E is a set with $n$ elements, $X \to E - X$ is a bijection of the set of subsets of E consisting of $p$ elements onto the set of subsets of $n - p$ elements.

We put $\binom{n}{p} = 0$ for each pair of natural integers such that $p > n$. With this convention the number of subsets of $p$ elements in a set of $n$ elements is $\binom{n}{p}$ for every natural integer $p$.