I draw numbers 1 through k (k≤10) out of a hat ten times at random, replacing the numbers after drawing them. If I disregard the case where I draw "1" all ten times, explain why the number of possible sequences is divisible by 11. (Result by a calculator is insufficient because anyone can do that easily.)
Now if I change the number '10' to another integer n in the above paragraph, can I still have a similar result; i.e., the total possible number of configurations is divisible by n+1? Does this work for all integers n? If so, prove it; if not, find all integers n it works for.
(In reply to
re(4): Replace '10' by n by Bractals)
The problem starts off with 10 draws out of [1,k] for any fixed
k<=10 (and > 0, presumably). This gives k^10 possiblities for the
sequence of outcomes of the 10 draws, but we are to discard the one
that is 1,1,1,1,1,1,1,1,1,1, which leaves us k^101 which for each
k=1,2,...,10 is divisible by 11. Now replace 10 by n, and replace 11 by
n+1.

Posted by Richard
on 20060905 01:24:36 