Question

Let be prime and$1\le k\le p$. Prove that divides the binomial coefficient $\left(\begin{array}{c}p\\ k\end{array}\right)$Recall that $\left[\left(\frac{p}{k}\right)=\frac{p!}{k!\left(p-k\right)!}\right]$

Short answer

Applying Theorem 1.4,$k!\left(p-k\right)!\left(p-1\right)!$and so$\frac{\left(p-1\right)!}{k!\left(p-k\right)!}$and hence, it is proved that the divides the binomial coefficient$p\left(\begin{array}{c}p\\ k\end{array}\right)$

Step-by-step solution

Define p

Consider the given equation

$\begin{array}{rcl}\left(\begin{array}{c}p\\ k\end{array}\right)& =& \frac{p!}{k!\left(p-k\right)!}\\ \left(\begin{array}{c}p\\ k\end{array}\right)& =& \frac{p!}{k!\left(p-k\right)!}\\ & =& p\frac{\left(p-1\right)!}{k!\left(p-k\right)!}\end{array}$

It is an integer, therefore $k!\left(p-k\right)!p!$

Here, $k<p$and $p-k<p$indicates that $\left(k!\left(p-k\right)!,p\right)=1$.

As $p$cannot appear in the prime decomposition of either$k!$or$\left(p-k\right)!$.

Using Theorem 1.4

Using Theorem 1.4,