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].