Even the extreme case where the last m items are chosen has the same probability; it requires the first n-m items not to be chosen, so the probability is:

((n-m)/n) * ((n-1-m)/(n-1)) * ((n-2-m)/(n-2)) * … * (1/(m + 1)) = (n-m)! / (n! / m!) = 1 / C(n,m)

as required. ]]>

BTW: +1 for a great and concise post!

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