exercise:4192d34079: Difference between revisions
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(Created page with "<div class="d-none"><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></div> 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 display="block"> {{{n + a - r...") |
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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 | |||
\ldots, n\}</math>, then the probability that an <math>a</math>-shuffle leads to an ordering with | |||
exactly <math>r</math> rising sequences equals | exactly <math>r</math> rising sequences equals | ||
Latest revision as of 00:29, 13 June 2024
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].