Question

give a combinatorial argument to prove the identity

k(n,k)=n(n-1,k-1)

I am looking at pascals identity right now to try and prove, but I am unsure how to actually work it except to explain the theorem ..... :(

Answer 1

R.H.S=n(n-1,k-1)

Answer 2

\[=n \left(\begin{matrix}n-1 \\k-1\end{matrix}\right)=n \frac{ \left( n-1 \right)! }{ \left\{ \left( n-1 \right)-\left( k-1 \right) \right\}!\left( n-1\right)!}\]

Answer 3

\[= \frac{n*\left( n-1 \right)! }{\left( n-k \right)!\left( k-1 \right)! }*\frac{ k }{k }\]

Answer 4

\[=\frac{ k \left( n! \right) }{\left( n-k \right)!*k! }=k \left(\begin{matrix}n \\ k\end{matrix}\right)\]
=L.H.S

Answer 5

THANK YOU!! i didn't even think to break it into its factorial.... i totally should have... but that helped ALOT!!!

Answer 6

YW