[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Show that if we start with the identity ordering of [math]\{1, 2, \ldots, n\}[/math], then the probability that an [math]a[/math]-shuffle leads to an ordering with

exactly [math]r[/math] rising sequences equals

[[math]] {{{n + a - r}\choose{n}}\over{a^n}}A(n, r)\ , [[/math]]

for [math]1 \le r \le a[/math].