Question

Prove the folliwing combinatorial identity if 1≤k 13 years ago

Answer 1

\[1\le k <n\]
\[\left(\begin{matrix}n-1 \\ k-1\end{matrix}\right) \left(\begin{matrix}n \\ k+1\end{matrix}\right) \left(\begin{matrix}n+1 \\ k\end{matrix}\right) = \left(\begin{matrix}n-1 \\ k\end{matrix}\right) \left(\begin{matrix}n \\ k-1\end{matrix}\right) \left(\begin{matrix}n+1 \\ k+1\end{matrix}\right)\]

Answer 2

What do you want us to find?

@zhengcl86

Answer 3

this question says
prove the folliwing combinatorial identity if 1≤k<n . this identity is known as the hexagon identity and relates terms in pascal's triangle

Answer 4

then its got that whole equation on the bottom (n−1k−1)(nk+1)(n+1k)=(n−1k)(nk−1)(n+1k+1)

Answer 5

This one is actually pretty easy if you just rewrite each combination as:
\[\left(\begin{matrix}a \\ b\end{matrix}\right)=\frac{ a! }{ (a-b)!b! }\].