1. ## Combinatorics

Show that:

$$\left( \begin{array}{cc} n \\ k \\ \end{array} \right)$$
$$\left( \begin{array}{cc} n-k \\ j \\ \end{array} \right)$$
= $$\left( \begin{array}{cc} n \\ j \\ \end{array} \right)$$
$$\left( \begin{array}{cc} n-j \\ k \\ \end{array} \right)$$

$j+k\leq n$

need some help here, only thing that came to my mind was that

$$\left( \begin{array}{cc} n \\ k \\ \end{array} \right)$$
= $$\left( \begin{array}{cc} n \\ n-k \\ \end{array} \right)$$

then what?

Thanks!

2. Hello, mechaniac!

$\displaystyle \text{Show that: } {n\choose k}{n-k\choose j} \;=\; {n\choose j}{n-j\choose k}$

$\displaystyle {n\choose k}{n\!-\!k\choose j} \;=\;\frac{n!}{k!\,(n\!-\!k)!}\cdot\frac{(n\!-\!k)!}{j!(n\!-\!j\!-\!k)!} \;=\;\frac{n!}{j!\,k!\,(n\!-\!j\!-\!k)!}$

$\text{Multiply by }\dfrac{(n\!-\!j)!}{(n\!-\!j)!}$

. . $\displaystyle \frac{(n\!-\!j)!}{(n\!-\!j)!}\cdot\frac{n!}{j!\,k!\,(n\!-\!j\!-\!k)!} \;=\;\frac{n!\,(n\!-\!j)!}{j!\,k!\,(n\!-\!j)!(n\!-\!j\!-\!k)!}$

. . $\displaystyle =\;\frac{n!}{j!(n\!-\!j)!} \cdot \frac{(n\!-\!j)!}{k!\,(n\!-\!j\!-\!k)!} \;=\;{n\choose j}{n\!-\!j\choose k}$

3. of course! thanks for the good explanation!