Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$\begin{aligned}\mathbf{D}_{p+1}^{n-1}\\\\\text{gedeeld door}\\\\\mathbf{D}_n^p\\\\\text{is gelijk aan}\end{aligned}$	A. $1 $
	B. $p $
	C. $n+1$
	D. $\frac {n\,+\,p} {p\,+\,1} $
	E. $\frac {n\,+\,p} {n} $

$\\\mathbf{D}_n^p=C_{n+p-1}^{\;p}=\;_{n+p+1}{\mathbf{C}}_{p}\:=\:^{n-p+1}{\mathbf{C}}_{p}=\left(_{\quad p}^{n+p-1}\right)=\mathbf{C}(n+p-1,p)$

[ 6-3834 - op net sinds 24.12.13-(E)-25.11.2024 ]

Translation in E N G L I S H

... divided by ... is equal to

Oplossing - Solution

$\mathbf{D_{\,n}^{\,p}}\,_{\;wordt\;ook\;geschreven\;als\;}^{\quad is\;\;also\;\;written\;\;as}\mathbf{\overline{C_{\,n}^{\,p}}$

$https://latex.codecogs.com/png.image?\dpi{110}\frac{D_{p+1}^{n-1}}{D_{n}^{p}}=\frac{C_{n+p-1}^{n-1}}{C_{n+p-1}^{p}}=1 \quad want \; (n\!-\!1)+p = n+p-1\\{\color{Green} .} \qquad \qquad \qquad \qquad\qquad(eigenschap\;\;C_{n}^{\,p}=C_{n}^{\,n-p} )$

$\begin{aligned}\mathbf{D}_{p+1}^{n-1}\\\\\text{gedeeld door}\\\\\mathbf{D}_n^p\\\\\text{is gelijk aan}\end{aligned}$	A. \(1 \)
	B. \(p \)
	C. \(n+1\)
	D. \(\frac {n\,+\,p} {p\,+\,1} \)
	E. \(\frac {n\,+\,p} {n} \)